I gather that some of these questions are tech active but I've completed fully worked solutions as it is quite likely you could get questions similar to this on non-tech related things. I feel it is important to know how these things work rather than blindly plugging values into your calculator (not to say this is something you would do
)
1. An investor received $1400 interest per annum from a sum of money, with part of it invested at 10% and the remainder at 7% simple interest. This investor found that if she interchanged the amounts she had invested she could increase her return by $90 per annum. Calculate the total amount invested.
I'm quite deficient when it comes to the wording of these problems, for I'm not entirely sure what some of the words mean (mainly because I'm lazy and don't want to learn simple things) so I won't have a go at this, but I'm sure someone else will help
2.Find the acute angle between each of the following straight lines:
a) with gradients 3 and 3/4
b) with equations 2y=8x+10 and 3x-6y=22
Note that
(see image below for understanding) where theta will be the angle between the line and the horizontal.
So to calculate the angle between two lines, we can calculate the angle between the steeper line then subtract the angle of the other to give the angle between, as follows;
)
)
Therefore
-tan^{-1}(3/4)=34.695 = 34.7 \ degrees)
b) Rearrange the two equations for y in terms of x:
Therefore m = 4
Therefore m =
)
)
Therefore
-tan^{-1}(\frac{1}{2})= 49.3987 = 49.4 \ degrees)
3. A cubic graph has the following features, a turning point at (3,0), a y-intercept at (0,18) and the graph passes through (1,16)
a)find the equation of the graph
b)f(x) is dilated by a factor of 2 from the x-axis, find the new equation
c) determine all the values for x if f(x)=10
a) A cubic graph will have the equation
=ax^{3}+bx^{2}+cx+d)
From this we know
=3ax^{2}+2bx+c)
Now we can insert the information given into the two equations
The graph goes through the point (3,0)
^{3}+b(3)^{2}+c(3)+d = 0)
The graph goes through the point (0,18)
^{3}+b(0)^{2}+c(0)+d = 18)
The graph goes through the point (1,16)
^{3}+b(1)^{2}+c(1)+d = 16)
Finally the graph has a turning point at (3,0) [so the derivative will be equal to zero at this point]
^{2} + 2b(3) + c = 0)
Now we can solve this on our calculator. One method is by matrices, as follows;
So this is in the form
So to solve it do
which will give us:
Thus,
 = 2x^{3}-10x^{2}+6x+18)
*Note the final part can be done by hand but it is immensely improbable you would be asked to do such a thing
b) A dilation by a factor of 2 from the x-axis, is the same as a dilation by a factor of 2 in the y-axis, so:
=2f(x)=2[2x^{3}-10x^{2}+6x+18]=4x^{3}-20x^{2}+12x+36)
c) Solve the original equation f(x) to equal 10;
= 2(4)^{3}-10(4)^{2}+6(4)+8=0)
Therefore (x - 4) is a factor
Solve by long division, as follows:
 & \:\:\underline{2x^{3}-8x^{2}}<br />\\ & \:\:\:\:\:\:\:\:\:\:\:-2x^{2}+6x<br />\\ (-) & \:\:\:\:\:\:\:\:\,\,\:\:\:\underline{-2x^{2}+8x}<br />\\ & \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:-2x-8<br />\\ (-) & \:\:\:\:\:\qquad\:\:\:\:\:\:\:\:\:\;\:\underline{-2x+8}<br />\\ & \qquad\qquad\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:0<br />\end{alignedat})
Now to find the other 2 solutions use the quadratic formula:
 \pm \sqrt{(-2)^{2}-4(2)(-2)}}{2(2)})
and from before also,
another function, g(x)=a/x-h+k, it has an x-intercept of (1,0) and asymptotes of x=-3 and y=1.
a) find the equation of g(x)
b) find the point of intersection between g(x) and f(X)
a)
 = \frac{a}{x+h}+k)
The vertical asymptote will occur when the denominator is equal to 0, thus:
The horizontal asymptote is given by the k value, hence
So now we have:
 = \frac{a}{x+3}+1)
We know it goes through the point (1,0), hence:
=\frac{-4}{x+3}+1)
The intersection of g(x) and f(x) will be where they have the same y-value, so g(x) = f(x):
Solve
on your calculator, as I doubt you would be able to get a real result by hand.
Hope I haven't made any mistakes and enjoy
edit: added latex'd long division - thanks to b^3 for that.
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