Post by: knightrider on January 07, 2014, 11:49:26 pm
a)Make f the subject of the equation
b)Make u the subject of the equation.
c)How far from the lens is the image when an object is 30 cm in front of a lens of focal length 25 cm?
2)The volume of a cone is given by the rule V=1/3pie*R^2*h, where r is the radius of the widest part of the cone and h is the vertical height of the cone. Given that the volume of a cone is 100 cm3 and its radius at the widest point is sqrt12 cm, find the height of the cone, expressing your answer in terms of pie.
Post by: RKTR on January 07, 2014, 11:59:16 pm
f=(uv)/(u+v)
b) 1/u = (v-f)/(fv)
u= (fv)/ (v-f)
c) 30= (25v)/(v-25)
30v-750=25v
5v=750
v=150cm
2. 100= (1/3)(pie)(12)h
100=4pie h
h= 25/ pie
Post by: knightrider on January 08, 2014, 12:29:53 am
can you help with this
The length of a side of a right-angled triangle can be found using Pythagoras' theorem: c2 = a2 + b2, where c is the length of the longest side, and a and b are the lengths of the two shorter sides. Find the value of b in the triangle above.
a=2root3
c=4
Post by: T-Infinite on January 08, 2014, 12:36:30 am
thankyou for your help
can you help with this
The length of a side of a right-angled triangle can be found using Pythagoras' theorem: c2 = a2 + b2, where c is the length of the longest side, and a and b are the lengths of the two shorter sides. Find the value of b in the triangle above.
a=2root3
c=4
Most of these questions just require you to sub in the given information.
So you're given, a and given c, just put it in the given equation:
Post by: Only Cheating Yourself on January 08, 2014, 11:31:45 am
1)The object and image positions for a lens of focal length f are related by the formula 1/u+1/v=1/f where u is the distance of the object from the lens and v is the distance of the image from the lens.
a)Make f the subject of the equation
b)Make u the subject of the equation.
c)How far from the lens is the image when an object is 30 cm in front of a lens of focal length 25 cm?
2)The volume of a cone is given by the rule V=1/3pie*R^2*h, where r is the radius of the widest part of the cone and h is the vertical height of the cone. Given that the volume of a cone is 100 cm3 and its radius at the widest point is sqrt12 cm, find the height of the cone, expressing your answer in terms of pie.
a.) I see f as a denominator and we need to be the subject. I see two fractions on the other side, i want to make f the numerator so i form a reporical but i can't do that because these two separate fractions so i need to make 1 fraction, so i cross multiply which is gong to make v+u as the numerator and so make the denominators same by multiplying so its v+u/vu=1/f now i can flip them so slip everything vu/v+u=f/1, now i have 1 as the denominator now to get rid of it i multiply both sides by 1 so its now 1 but you don't have to write it so vu/v+u=f and now F is the subject
b.)So we want u the subject, so let 1/u stay as it is and we'll move 1/v to the right side so, 1/u=1/f- 1/v, now we have 1/u by its self we need to make u the numerator as its easier to make it the subject, so again we cross multiply so we can combine the fractions into 1, so 1/u=v-f/vf now we flip them to make u the numerator so u/1=v-f/vf, so get rid of the 1 we multiply both sides but again we don't need to write the 1 as 1 times anything just stays the same so the answer is u=v-f/vf
c.) I had trouble with this one as well! So the questions asking how far from the lens is the image which is what v stands for so we need to make v the subject. So get 1/v on one side so move 1/u to the other side so
1/v=1/f-1/u now cross multiply to make it 1 fraction so 1/v=u-f/fu, now flip it to make v the numerator so flip both sides v/1=fu/u-f , multiply 1 to both sides so now we have made v the subject. v=fu/u-f, now plug in the information provided so 30*25/30-25=150cm the answer is 150cm
Post by: knightrider on January 08, 2014, 09:00:38 pm
For safety considerations, wheelchair
ramps are constructed under regulated
specifications. One regulation requires
that the maximum gradient of a ramp
exceeding 1200 mm in length is 1/14
a)Does a ramp 25 cm high with a
horizontal length of 210 cm meet
the requirements?
b)Does a ramp with gradient 1/18 meet the specifications?
c)A 16 cm high ramp needs to be
built. Find the horizontal length
of the ramp required to meet the
specifications.
Post by: Phy124 on January 08, 2014, 09:20:31 pm
thx what about this questionThe length is 210cm or 2100mm therefore it exceeds a length of 1200mm and hence needs a gradient less than 1/14.
For safety considerations, wheelchair
ramps are constructed under regulated
specifications. One regulation requires
that the maximum gradient of a ramp
exceeding 1200 mm in length is 1/14
a)Does a ramp 25 cm high with a
horizontal length of 210 cm meet
the requirements?
The gradient is given by the rise/run which is 25/210 = 5/42
1/14 is equal to 3/42 so we know it does not meet the requirements as 5/42 > 3/42
b)Does a ramp with gradient 1/18 meet the specifications?1/18 < 1/14, so yes.
c)A 16 cm high ramp needs to beIf the horizontal length is less than 1200mm then the gradient could be anything, so we can just assume to start with the horizontal length will be greater than 1200mm and hence is restricted by a gradient of 1/14.
built. Find the horizontal length
of the ramp required to meet the
specifications.
rise/run = 16/horizontal length = 1/14
Therefore the horizontal length is equal to 14*16 = 224cm
Post by: knightrider on January 08, 2014, 09:29:30 pm
Which of the following lines has a
gradient of −2?
Which of the following lines has a
gradient of 3?
how would i do this?
Post by: Phy124 on January 08, 2014, 09:39:53 pm
If a line goes right and upwards it has a positive rise and a positive run +/+ = +, hence positive gradient.
If a line goes left and upwards it has a positive rise and a negative run +/- = -, hence negative gradient.
If a line goes right and downwards it has a negative rise and a positive run -/+ = -, hence negative gradient.
If a line goes left and downwards it has a negative rise and a negative run -/- = +, hence positive gradient.
The first one (6a) is easy because you are looking for a line with a negative gradient and there is only one line (B) which has this property.
A and D have positive gradients, E has a 0 gradient and C has an undefined gradient.
For the second one (6b) we are looking for a line with positive gradient, this means it can be on of B, C, D and E (A has an undefined gradient).
To find the value of the gradient take two points (x1, y1) and (x2, y2) and use the formula:
You'll find that if you look at line E and take the points (x1, y1) = (4,1) and (x2, y2) = (2,-5) that it has gradient 3.
Post by: knightrider on January 09, 2014, 04:35:13 pm
5)The sum of two whole numbers, x and y, is 41. The difference between them is 3. Write two equations involving x and y and solve them to find the numbers.
6)A farmer counts emus and sheep in a paddock, and notes there are 57 animals and 196 feet. Assuming no animal amputees, how many of each animal are there?
7)A sports store supplies 24 basketballs and 16 cricket balls to one school for $275.60, and delivers 12 basketballs and 32 cricket balls to another school for $211. If delivery is free, how much did the supplier charge for each type of ball?
8)A businessperson hires a stretch limousine for 2 days and a sedan for 3 days while on an interstate trip. If the total car hire cost of the trip was $675, and the limousine cost triple the price of the sedan, find the cost per day of the limousine.
9)this is mulitple choice
A manufacturing plant produces square and circular metal panels in fixed sizes. If the mass of a square panel is 13 kg and that of a circular panel is 22 kg, how many of each panel are there in a truck loaded with 65 panels of total mass 1205 kg? The equations to solve are:
a)13s + 22c = 1205, s + c = 65
b)22s + 13c = 1205, s + c = 65
c)13s + 22c = 65, s + c = 1205
d)22s + 13c = 65, s + c = 1205
e)13s + 22c = 1205, s + c = 35
plz show full working out for questions thx
Post by: Phy124 on January 09, 2014, 05:05:37 pm
what about these questions using simultaneous equationsLet x equal the higher number and y equal the lower number.
5)The sum of two whole numbers, x and y, is 41. The difference between them is 3. Write two equations involving x and y and solve them to find the numbers.
You can derive one equation by noting that the sum of the two numbers is 41 and another by noting that the difference (higher number minus lower number is 3)
Spoiler
6)A farmer counts emus and sheep in a paddock, and notes there are 57 animals and 196 feet. Assuming no animal amputees, how many of each animal are there?Let x = emus and y = sheep
Emus have 2 feet and sheep have 4.
The total amount of animals will be the sum of the emus (x) and sheep (y) and this is equal to 57, this is our first equation.
Next, total number of feet is the sum of the number of emus (x) multiplied by the number of feet on each emu (2) and the number of sheep (y) multiplied by the number of feet on each sheep (4), this is equal to 196, so we have our second equation.
Spoiler
Therefore we have 16 emus and 41 sheep.
7)A sports store supplies 24 basketballs and 16 cricket balls to one school for $275.60, and delivers 12 basketballs and 32 cricket balls to another school for $211. If delivery is free, how much did the supplier charge for each type of ball?Let x = price of basketball and y = price of cricket ball
The store supplies 24 basketballs at $x and 16 cricket balls at $y for a total price of $275.60, this forms our first equation.
The store then supplies 12 basketballs at $x and 32 cricket balls at $y for a total price of $211, this forms our second equation.
Spoiler
Therefore basketballs cost $9.45 and cricket balls cost $3.05[/tex]
I'll leave the last two for you to do but I'll give you some hints.
8)A businessperson hires a stretch limousine for 2 days and a sedan for 3 days while on an interstate trip. If the total car hire cost of the trip was $675, and the limousine cost triple the price of the sedan, find the cost per day of the limousine.
Spoiler
Let x = the cost of the limousine per day and y = the cost of the sedan per day.
You are looking for the cost per day of the vehicles and thus you need to form two equations relating the number of days each vehicle was hired and the amount it costs per day. (e.g. if a vehicle clost $z per day and was hired for 5 days it would cost a total price of 5z over the 5 days).
Do this for both vehicles and sum them to equal the total price.
Note that you also have a relationship that the limousine costs three times as much as the sedan, therefore x = 3y, this is your second equation.
You are looking for the cost per day of the vehicles and thus you need to form two equations relating the number of days each vehicle was hired and the amount it costs per day. (e.g. if a vehicle clost $z per day and was hired for 5 days it would cost a total price of 5z over the 5 days).
Do this for both vehicles and sum them to equal the total price.
Note that you also have a relationship that the limousine costs three times as much as the sedan, therefore x = 3y, this is your second equation.
9)this is mulitple choice
A manufacturing plant produces square and circular metal panels in fixed sizes. If the mass of a square panel is 13 kg and that of a circular panel is 22 kg, how many of each panel are there in a truck loaded with 65 panels of total mass 1205 kg? The equations to solve are:
a)13s + 22c = 1205, s + c = 65
b)22s + 13c = 1205, s + c = 65
c)13s + 22c = 65, s + c = 1205
d)22s + 13c = 65, s + c = 1205
e)13s + 22c = 1205, s + c = 35
plz show full working out for questions thx
Spoiler
Let s = the number of square panels and c = the number of circular panels
The total weight will be given by the number of square panels (s) multiplied by the weight of a square panel (13) added to the number of circular panels (c) multiplied by the weight of a circular panel (22), this weight is given by 1205. This forms your first equation.
The second equation is simply the number of square panels (s) added to the number of circular panels (c) to equal the number of total panels (65).
The total weight will be given by the number of square panels (s) multiplied by the weight of a square panel (13) added to the number of circular panels (c) multiplied by the weight of a circular panel (22), this weight is given by 1205. This forms your first equation.
The second equation is simply the number of square panels (s) added to the number of circular panels (c) to equal the number of total panels (65).
Post by: Butterscotch on January 09, 2014, 05:09:51 pm
what about these questions using simultaneous equations
5)The sum of two whole numbers, x and y, is 41. The difference between them is 3. Write two equations involving x and y and solve them to find the numbers.
Picture attached:
Edit: Beaten by Phy124.
Post by: knightrider on January 09, 2014, 07:30:25 pm
Post by: LOLs99 on January 09, 2014, 07:49:55 pm
how would you find the equations for these graphs
Hint: use the general equation for linear y= mx+c and sub in coordinates/x or y int.
Post by: Phy124 on January 09, 2014, 07:50:23 pm
how would you find the equations for these graphsBecause you are shown the y-intercept it is easiest to put it straight in the form
Let's do question a.
We can see from the graph that the y-intercept is 2 and hence
Now to find the gradient we take any two points. I'll use
Post by: Only Cheating Yourself on January 09, 2014, 08:11:38 pm
how would you find the equations for these graphs
So each line goes through the x and y axis. For e.g the first one a.) goes through the x and y axis, x axis (-1/2,0) y axis (0,2) nah you have your two coordinates use the formula to find the gradient, 2-0=2, 0--1/2=1/2 so 2/1/-1/2=4 so gradient is 4. Now we can find the equation.
(0,2)
y-2=4(x-0)
y-2=4x
-2=4x-y
4x-y+2=0
-y=-2
y=2
put it in mx+c form
y=4x+2
Post by: knightrider on January 13, 2014, 02:32:19 pm
1)Find the equation of the line passing through (3, −3) that makes an angle of 45° with the positive x-axis.
2)Find the equation of the line containing (7, −2) that makes an angle of 71.565° with the positive x-axis.
3)Find the equation of the line (in y = mx + c form) that:
a)is perpendicular to the line with equation y = 3x + 1, passing through (−3, 6)
b)is parallel to the line with equation y = 2/5x-9, passing through (4, −7)
c)is parallel to the line with equation 3x + 6y = 8, passing through (2, 2)
d)is perpendicular to the line with equation −6x + 7y − 2 = 0, passing through (4, 0)
e)has gradient 2, passing through the intersection of the lines with equations y = 3x − 5 and y = −2x + 5
f)has gradient -3/4, passing through the intersection of the lines with equations x + 4y = −14 and −5x + 2y = 4.
4)Find the equation of the line that passes through the point of intersection of the lines whose equations are 7x − 3y − 19 = 0 and 3x + 2y + 5 = 0, given that the required line is parallel to the line with equation −5x − 2y = 3.
Post by: Phy124 on January 13, 2014, 02:35:40 pm
Post by: Only Cheating Yourself on January 13, 2014, 02:52:39 pm
what about these questions
1)Find the equation of the line passing through (3, −3) that makes an angle of 45° with the positive x-axis.
2)Find the equation of the line containing (7, −2) that makes an angle of 71.565° with the positive x-axis.
3)Find the equation of the line (in y = mx + c form) that:
a)is perpendicular to the line with equation y = 3x + 1, passing through (−3, 6)
b)is parallel to the line with equation y = 2/5x-9, passing through (4, −7)
c)is parallel to the line with equation 3x + 6y = 8, passing through (2, 2)
d)is perpendicular to the line with equation −6x + 7y − 2 = 0, passing through (4, 0)
e)has gradient 2, passing through the intersection of the lines with equations y = 3x − 5 and y = −2x + 5
f)has gradient -3/4, passing through the intersection of the lines with equations x + 4y = −14 and −5x + 2y = 4.
4)Find the equation of the line that passes through the point of intersection of the lines whose equations are 7x − 3y − 19 = 0 and 3x + 2y + 5 = 0, given that the required line is parallel to the line with equation −5x − 2y = 3.
Yea what PHY said… Remember perpendicular is equal to -1.
Have an attempt at the questions first and show your working/thought process and then I'll lead you in the right direction :)
Hey, i won't answer the questions above… But i know how do them but i don't why i do them if that makes sense…. Im doing meth 1-2 to as well, but sometimes i do the working out and get the correct answer but i don't get why i do it. For if i wanted to the equation of a line perpendicular …. Forget its to hard to explain.
Post by: Daenerys Targaryen on January 13, 2014, 03:19:54 pm
Yea what PHY said… Remember perpendicular is equal to -1.
Hey, i won't answer the questions above… But i know how do them but i don't why i do them if that makes sense…. Im doing meth 1-2 to as well, but sometimes i do the working out and get the correct answer but i don't get why i do it. For if i wanted to the equation of a line perpendicular …. Forget its to hard to explain.
If you're referring to the gradient of a line perpendicular to another, the perpendicular line holds a gradient of the negative reciprocal of the gradient of the line it is perpendicular to.
Post by: Only Cheating Yourself on January 13, 2014, 03:28:00 pm
If you're referring to the gradient of a line perpendicular to another, the perpendicular line holds a gradient of the negative reciprocal of the gradient of the line it is perpendicular to.
Yea I'm confused.
Post by: Snorlax on January 13, 2014, 03:49:44 pm
Yea I'm confused.Say the gradient of the normal was
Then the gradient of the line perpendicular to it would be
Post by: hobbitle on January 13, 2014, 03:52:00 pm
Have an attempt at the questions first and show your working/thought process and then I'll lead you in the right direction :)
This.
Don't just regurgitate questions from a textbook at us. If you don't show working, we can't figure out where you're going wrong, and you don't actually learn anything.
Also, a 'thanks' after someone writes you pages and pages of working would be nice... The people who have been helping you on this thread have been too kind already.
Post by: Only Cheating Yourself on January 13, 2014, 05:00:12 pm
Say the gradient of the normal was.
Then the gradient of the line perpendicular to it would be
Yea i know but its kind of because I've done it so many times, if you get what i mean? For e.g if a q says find the equation of a line perpendicular to this equation y=4x+1 i straight away know what to do but its not because i understand it though.. the gradient for line perpendicular to that equation would be -1/4 as that =-1 so now thats the gradient of the equation of the line we're trying to find etc.
Also who keeps thumbing me down?
Post by: hobbitle on January 14, 2014, 10:34:24 am
the gradient for line perpendicular to that equation would be -1/4 as that =-1
I've been holding off saying this for a while. But when asking questions on this forum it's important to be clear so we can understand what you're asking. Please please please re-read your posts and ensure they make sense. To me, it reads like you just wrote -1/4 = -1 which clearly you know it doesn't so you must have meant something else. Almost all of your posts contain a lot of typos and grammatical errors that make interpreting a question very difficult. Please make a little more effort for the people who are putting in a lot of effort for you.
(No, I didn't down vote you).
Post by: Only Cheating Yourself on January 14, 2014, 10:39:37 am
I've been holding off saying this for a while. But when asking questions on this forum it's important to be clear so we can understand what you're asking. Please please please re-read your posts and ensure they make sense. To me, it reads like you just wrote -1/4 = -1 which clearly you know it doesn't so you must have meant something else. Almost all of your posts contain a lot of typos and grammatical errors that make interpreting a question very difficult. Please make a little more effort for the people who are putting in a lot of effort for you.
(No, I didn't down vote you).
Ok, i'll grab the dictionary out before i post things.
Post by: hobbitle on January 14, 2014, 10:41:08 am
Ok, i'll grab the dictionary out before i post things.
Sarcasm not required, cheers.
Post by: grannysmith on January 14, 2014, 10:58:40 am
Ok, i'll grab the dictionary out before i post things.Lol I doubt a dictionary would help.
But basically, what you're trying to say is that the product of two gradients that make the lines perpendicular to each other is -1.
So you've got a gradient of 2; what is the gradient of the line perpendicular to it?
2 x m = -1
Therefore m = -1/2
Post by: Only Cheating Yourself on January 14, 2014, 12:20:03 pm
Lol I doubt a dictionary would help.
But basically, what you're trying to say is that the product of two gradients that make the lines perpendicular to each other is -1.
So you've got a gradient of 2; what is the gradient of the line perpendicular to it?
2 x m = -1
Therefore m = -1/2
Whats that suppose to mean? And ye.
Post by: knightrider on January 15, 2014, 11:21:36 pm
what about these questions
1)Find the equation of the line passing through (3, −3) that makes an angle of 45° with the positive x-axis.
2)Find the equation of the line containing (7, −2) that makes an angle of 71.565° with the positive x-axis.
3)Find the equation of the line (in y = mx + c form) that:
a)is perpendicular to the line with equation y = 3x + 1, passing through (−3, 6)
b)is parallel to the line with equation y = 2/5x-9, passing through (4, −7)
c)is parallel to the line with equation 3x + 6y = 8, passing through (2, 2)
d)is perpendicular to the line with equation −6x + 7y − 2 = 0, passing through (4, 0)
e)has gradient 2, passing through the intersection of the lines with equations y = 3x − 5 and y = −2x + 5
f)has gradient -3/4, passing through the intersection of the lines with equations x + 4y = −14 and −5x + 2y = 4.
4)Find the equation of the line that passes through the point of intersection of the lines whose equations are 7x − 3y − 19 = 0 and 3x + 2y + 5 = 0, given that the required line is parallel to the line with equation −5x − 2y = 3.
for 3f i got the answer as y=-3/4x+55/18 but in the answers it says the answer is y=-3/4x-9/2 is the answer wrong or am i wrong ?
Post by: Special At Specialist on January 16, 2014, 12:03:54 am
for 3f i got the answer as y=-3/4x+55/18 but in the answers it says the answer is y=-3/4x-9/2 is the answer wrong or am i wrong ?
The latter. I'll show you the solution:
"Find the equation of the line that has gradient -3/4, passing through the intersection of the lines with equations x + 4y = −14 and −5x + 2y = 4."
y = mx + c
Since the gradient is -3/4, then m = -3/4
y = (-3/4)x + c
Now in order to work out c, you need to substitute the point (x, y) in. They have told you that the point is where the equations x + 4y = -14 and -5x + 2y = 4 intersect, so use this information to work out the point:
x + 4y = -14 ---> Equation 1
-5x + 2y = 4 ---> Equation 2
Solve the simultaneous equations:
From equation 1:
x = -14 - 4y
x = -(4y + 14)
Substitute into equation 2:
-5(-(4y + 14)) + 2y = 4
5(4y + 14) + 2y = 4
20y + 70 + 2y = 4
22y = -66
y = -3
Substitute into equation 1:
x = -14 - 4(-3)
x = -14 + 12
x = -2
So the point (x, y) = (-2, -3). Substitute this into your original equation:
y = (-3/4)x + c
If x = -2 and y = -3
-3 = (-3/4)(-2) + c
-3 = 3/2 + c
c = -3 - 3/2
c = -9/2
So the equation is:
y = (-3/4)x - 9/2
Post by: knightrider on January 16, 2014, 04:39:50 pm
how would you do these questions i cant be able to figure them out
1)A line passes through the points (−8, −5), (4, −3) and (a, 12). Find the value of a.
2)The points (2, 7) and (6, 9) lie on the same straight line. Does the point (4,8) also lie on this line?
Post by: Nato on January 16, 2014, 04:46:24 pm
thx for help
how would you do these questions i cant be able to figure them out
1)A line passes through the points (−8, −5), (4, −3) and (a, 12). Find the value of a.
2)The points (2, 7) and (6, 9) lie on the same straight line. Does the point (4,8) also lie on this line?
for question 1, you know two points for sure: (-8,-5), (4,-3). because of this you can actually find the equation of the line.
After finding the equation of the line you can sub in the point (a,12) to find the value of a. Do you get this?
and it's a similar process of 2). Find the equation of the line using the first two point. then you have the point (4,8). to test if this point lies on the line, sub it in to your new found equation. if you sub in x=4 and get y=8, and sub y=8 and get x=4. Then the point lies on the line.
Post by: Sup on January 16, 2014, 04:54:10 pm
Post by: Phy124 on January 16, 2014, 05:07:29 pm
thx for helpAn altnernative approach to that of Nato would be that the gradient between each of the points must be the same i.e.
how would you do these questions i cant be able to figure them out
1)A line passes through the points (−8, −5), (4, −3) and (a, 12). Find the value of a.
Knowing this we can equate 2 of these (if 2 are the same, then all 3 will be) to find the value of
2)The points (2, 7) and (6, 9) lie on the same straight line. Does the point (4,8) also lie on this line?
The same again here, check whether 2 of the following hold:
Therefore it does lie on the line.
*Note: You could also just do this by inspection. The points (2,7),(4,8),(6,9) obviously follow a linear trend.
Edit:
Can someone help me with this question? Thanks a lotYou have three points on the parabola.
1. The left most point: (0,75)
2. The local minimum: (90,30)
3. The right most point (following symmetry): (180,75)
You can put these into the equation
Alternatively you could have it in the form
Post by: Only Cheating Yourself on January 16, 2014, 05:19:18 pm
Can someone help me with this question? Thanks a lot
Hey ill do my best, but I'm a bit stuck as well and maybe if someone could tell me if I'm one the right track?
Anyway if we want to describe this graph, we can put it into turning point form a(x-b)^2+c
so the minimum is 30 as it states hence c=30, now we need to find b and the dilation a
it says that assume the parabola is the suspension cable but we can see the parabola does not touch not cross through the x intercepts meaning the discriminant is 1>discriminant so therefore it has no solutions so it has no x intercepts.
So now we know we have no x intercepts out turning point is (0,30) am i on the right track here?
Ok so now we have the turning pint, a and c we need to find A the dilation and this is where I'm stuck.
Tried my best.
Post by: Only Cheating Yourself on January 16, 2014, 05:38:32 pm
thx for help
how would you do these questions i cant be able to figure them out
1)A line passes through the points (−8, −5), (4, −3) and (a, 12). Find the value of a.
2)The points (2, 7) and (6, 9) lie on the same straight line. Does the point (4,8) also lie on this line?
1.) Those coordinates are along 1 line, therefore have the same gradient. So find the gradient of (-8,-5) (4,-3)=1/6
so gradient is 1/6. Now get coordinates (-8,-5) and (a,12) now find the gradient of this and it has to =1/6,
so 12+5=17, so 17 divided by something needs to give us 1/6, so we find that 17 times 6=102.
now 17/102=1/6 so it works! Not we have to find a so some thing a+8=102, so a=94, because 94+8=102
a=94
2.) yes they do because (2,7) and (6,9) =1/2 so their gradient is 1/2 and if you find the gradient of (4,8) with (2,7) you'll find that it =1/2.
Post by: LOLs99 on January 16, 2014, 05:39:32 pm
Hey ill do my best, but I'm a bit stuck as well and maybe if someone could tell me if I'm one the right track?
Anyway if we want to describe this graph, we can put it into turning point form a(x-b)^2+c
so the minimum is 30 as it states hence c=30, now we need to find b and the dilation a
it says that assume the parabola is the suspension cable but we can see the parabola does not touch not cross through the x intercepts meaning the discriminant is 1>discriminant so therefore it has no solutions so it has no x intercepts.
So now we know we have no x intercepts out turning point is (0,30) am i on the right track here?
Ok so now we have the turning pint, a and c we need to find A the dilation and this is where I'm stuck.
Tried my best.
Why is C=30?
Post by: Only Cheating Yourself on January 16, 2014, 05:49:17 pm
Why is C=30?
Maximum and minimum is given by c in the form of a turning pint form hence c=30.
Post by: hobbitle on January 16, 2014, 05:53:48 pm
So now we know we have no x intercepts out turning point is (0,30) am i on the right track here?
The numbers are a little awkward I think and I've kind of forgotten turning point form but I think you went wrong about half way.
You have three known points from the information given that lie on the graph:
(0,75) - the extreme left of the parabola
(90,30) - the turning point... the 90 comes from half way between the 180m distance provided, the 30 is stated in the question.
(180,75) - the extreme right of the parabola
turning point form:
y = a(x-h)^2 + c
so we can sub in our turning point:
y = a(x - 90)^2 + 30
then we can sub in one of our other points to find 'a', lets use (0,75)
75 = a(0 - 90)^2 + 30
expand
75 = 8100a + 30
subtract 30 from both sides
45 = 8100a
divide both sides by 8100
a = 45/8100
so your final answer in turning point form is
y = 45/8100 (x - 90)^2 + 30
Post by: Only Cheating Yourself on January 16, 2014, 05:55:38 pm
An altnernative approach to that of Nato would be that the gradient between each of the points must be the same i.e.
Knowing this we can equate 2 of these (if 2 are the same, then all 3 will be) to find the value of
The same again here, check whether 2 of the following hold:
Therefore it does lie on the line.
*Note: You could also just do this by inspection. The points (2,7),(4,8),(6,9) obviously follow a linear trend.
Edit:
You have three points on the parabola.
1. The left most point: (0,75)
2. The local minimum: (90,30)
3. The right most point (following symmetry): (180,75)
You can put these into the equationand solve for
.
Alternatively you could have it in the formand sub in one of the end points.
Thought i was on the right track… Then saw this.
Post by: Only Cheating Yourself on January 16, 2014, 05:58:32 pm
The numbers are a little awkward I think and I've kind of forgotten turning point form but I think you went wrong about half way.
You have three known points from the information given that lie on the graph:
(0,75) - the extreme left of the parabola
(90,30) - the turning point... the 90 comes from half way between the 180m distance provided, the 30 is stated in the question.
(180,75) - the extreme right of the parabola
turning point form:
y = a(x-h)^2 + c
so we can sub in our turning point:
y = a(x - 90)^2 + 30
then we can sub in one of our other points to find 'a', lets use (0,75)
75 = a(0 - 90)^2 + 30
expand
75 = 8100a + 30
subtract 30 from both sides
45 = 8100a
divide both sides by 8100
a = 45/8100
so your final answer in turning point form is
y = 45/8100 (x - 90)^2 + 30
Didn't see that! Yep turning point of x is half way of the x intercepts… Should have saw that. But the parabola doesn't touch the x intercepts so i thought it had no x intercepts.
Post by: hobbitle on January 16, 2014, 05:59:31 pm
Didn't see that! Yep turning point of x is half way of the x intercepts… Should have saw that. But the parabola doesn't touch the x intercepts so i thought it had no x intercepts.
You're right, it doesn't have any x intercepts.
That doesn't mean that the x-coordinate of the turning point is zero, though. They are unrelated things.
The extremities of the parabola along the x axis (0 --> 180) aren't intercepts. They are just points along the x axis. They aren't 'cutting' the x axis.
Post by: Only Cheating Yourself on January 16, 2014, 06:01:23 pm
You're right, it doesn't have any x intercepts.
That doesn't mean that the x-coordinate of the turning point is zero, though. They are unrelated things.
but to find the x coordinate of the turning point don't you add the two x intercepts together, if there isn't any 0+0/2=0 so isn't it still 0?
Post by: Daenerys Targaryen on January 16, 2014, 06:03:24 pm
but to find the x coordinate of the turning point don't you add the two x intercepts together, if there isn't any 0+0/2=0 so isn't it still 0?If there aren't any intercepts, it doesn't naturally follow that the value is 0. The value being zero means the intercept is at 0. So the x int is undefined as there is none
Post by: hobbitle on January 16, 2014, 06:04:11 pm
You're looking at the half way point between the two extremities along the x axis. It has nothing to do with an x intercept.
Post by: Only Cheating Yourself on January 16, 2014, 06:07:52 pm
...no.... not all graphs have an x intercept..... so wherever you got that idea/formula from, it's wrong and useless most of the time.
You're looking at the half way point between the two extremities along the x axis. It has nothing to do with an x intercept.
My textbook shows that to find the turning point of x you add the two x intercepts the divide by 2. Im confused as hell! So if I'm given an application question and it says the x axis is 200m long does this mean the x coordinate of the turning point is 100? Then what the hell am i talking about? I've been answering questions from my book to find the turning point by adding the x intercepts then dividing by 2 so is the turning point value of always half the x axis? So is it the x axis is 250 the turning point of x is 125?
Post by: Daenerys Targaryen on January 16, 2014, 06:11:54 pm
My textbook shows that to find the turning point of x you add the two x intercepts the divide by 2. Im confused as hell! So if I'm given an application question and it says the x axis is 200m long does this mean the x coordinate of the turning point is 100? Then what the hell am i talking about? I've been answering questions from my book to find the turning point by adding the x intercepts then dividing by 2 so is the turning point value of always half the x axis? So is it the x axis is 250 the turning point of x is 125?This only works for parabolas
But a parabola does not always have intercepts
They can have 2, 1 or none
One int is when the parabola turns (has a TP) on the x axis
No ints is when the TP is above the x axis and is a positive parabola OR the TP is below the x axis and a negative parabola.
I reccommend sketching a little graph of your eqn to see if interecepts exist, or use the discriminant formula
EDIT: The domain of the graph could also be restricted, so that the graph can only be showing parts of a graph (and omit the x intercepts)
Post by: Phy124 on January 16, 2014, 06:13:28 pm
My textbook shows that to find the turning point of x you add the two x intercepts the divide by 2. Im confused as hell! So if I'm given an application question and it says the x axis is 200m long does this mean the x coordinate of the turning point is 100? Then what the hell am i talking about? I've been answering questions from my book to find the turning point by adding the x intercepts then dividing by 2 so is the turning point value of always half the x axis? So is it the x axis is 250 the turning point of x is 125?
The x value for the turning point on a parabola will be halfway between two points that share the same y value, as per symmetry.
In the case you speak about the two x-intercepts share the same y value (0) and hence the x value for the turning point is halfway between them.
Post by: Only Cheating Yourself on January 16, 2014, 06:13:38 pm
This only works for parabolas
But a parabola does not always have intercepts
They can have 2, 1 or none
One int is when the parabola turns (has a TP) on the x axis
No ints is when the TP is above the x axis and is a positive parabola OR the TP is below the x axis and a negative parabola.
I reccommend sketching a little graph of your eqn to see if interecepts exist, or use the discriminant formula
But the questions says its a parabola and you said this only works for parobalas wasn't the question a parabola graph?
The x value for the turning point on a parabola will be halfway between two points that share the same y value, as per symmetry.
In the case you speak about the two x-intercepts share the same y value (0) and hence the x value for the turning point is halfway between them.
So then how did you guys get 90? When there are no x intercepts?
Post by: hobbitle on January 16, 2014, 06:13:59 pm
I guess the simplest way is to think of it like symmetry. Just use intuition, don't just 'do what the textbook tells you'. Actually look at the picture and think about it. You know that that graph provided is symmetrical... the leftmost point is at 75 (on the y axis) and the rightmost point is at 75 (on the y axis). Because the curve of a parabola is by definition symmetrical, you know that the turning point must be in the middle....
Post by: LOLs99 on January 16, 2014, 06:14:38 pm
Maximum and minimum is given by c in the form of a turning pint form hence c=30.Oh c = 30 is right too. I went for y=ax^2+bx+c formula so my c is 75. Sorry about that :)
Post by: knightrider on January 16, 2014, 06:43:28 pm
how do you go about answering these types of questions?
i need help with this question
Post by: hobbitle on January 16, 2014, 06:47:45 pm
thankyou guys for your help
how do you go about answering these types of questions?
i need help with this question
I'm not spoon-feeding you this one until you show us some working or at least express some thought patterns to us (and I hope none else just tells you the answer either).
Key points: "linear manner". The graph is a line. So you know the format of the graph, y=mx+c.
You also have been given two coordinates on said graph. They are just given to you in the form of pictures.
Use the two coordinates to find the equation of the line (I believe you know how to do that already).
Post by: b^3 on January 16, 2014, 06:49:26 pm
We will also let
Now we can note two points, we know that after two months (
Now you have two points, you'd find the equation as you would normally with a straight line. i.e. Find the gradient
Post by: Sup on January 16, 2014, 07:00:08 pm
An altnernative approach to that of Nato would be that the gradient between each of the points must be the same i.e.
Knowing this we can equate 2 of these (if 2 are the same, then all 3 will be) to find the value of
The same again here, check whether 2 of the following hold:
Therefore it does lie on the line.
*Note: You could also just do this by inspection. The points (2,7),(4,8),(6,9) obviously follow a linear trend.
Edit:
You have three points on the parabola.
1. The left most point: (0,75)
2. The local minimum: (90,30)
3. The right most point (following symmetry): (180,75)
You can put these into the equation
Alternatively you could have it in the form
Thanks, could you please put it in general form. Thanks
Post by: Phy124 on January 16, 2014, 07:17:30 pm
Thanks, could you please put it in general form. ThanksNot sure what you mean but:
Where
edit:
I see what you mean
Post by: Sup on January 16, 2014, 07:19:49 pm
Not sure what you mean but:
Whereis the coordinate of the turning point and
is a dilation factor.
I know that in turning point form, it is: y=1/180(x-90)^2 +30, but what is this in the form of y = ax^2 +bx + c?
Thanks
Post by: hobbitle on January 16, 2014, 07:23:44 pm
Post by: Sup on January 16, 2014, 07:25:50 pm
Post by: Sup on January 16, 2014, 08:16:30 pm
Thanks.
Post by: Daenerys Targaryen on January 16, 2014, 08:21:01 pm
Got another question: A parabola has the same shape as y=2x^2 but its turning point is (1, -2). Write its equation.
Thanks.
The 'a' in the TP form: a(x-b)^2 + c
is usually the 'a' in the basic form, ax^2 + bx +c like it is in this case
Thus given the TP, we sub in b and c so that it becomes:
y=2(x-1)^2 -2