Author Topic: need help with these questions (Read 1166 times) Tweet Share

huss48 · « **on:** March 10, 2010, 04:26:58 pm »

I'm having trouble in doing these questions and can any1 work it out and explain how get the answer . THANKS

1. If two lines 3x-y+4=0 and ax+2y-3=0 are perpendicular then
a=2/3 a=-1/3 a=-2/3 a= 1/3 a=1/6

2. if |p^2-5p|=6 then p =
-1 6 -1,6 2,3 -1,2,3,6

3.Consider the function with wquation y= 4/(x-2)^2 -1
A. Algebraically find y-intercept & x intercept

B. State the equation of the asymptotes

Thanks you for ur help

TrueTears · « **Reply #1 on:** March 10, 2010, 04:33:11 pm »

2. p|p-5|=6

from here we see p>0 so it is either 6 or 2,3

the.watchman · « **Reply #2 on:** March 10, 2010, 04:34:44 pm »

1) $3x-y+4=0$ , $y=3x+4$

So the perpendicular gradient is $m=\frac{-1}{3} = \frac{-a}{2}$ , $a=\frac{2}{3}$

m@tty · « **Reply #3 on:** March 10, 2010, 04:38:48 pm »

Quote from: TrueTears on March 10, 2010, 04:33:11 pm

2. p|p-5|=6

from here we see p>0 so it is either 6 or 2,3

Are you being serious? I assume you're not...

$\begin{align}|p^2-5p|&\ne p|p-5| \\ &=|p||p-5| \end{align}$

-1 is also a solution.

huss48 · « **Reply #4 on:** March 10, 2010, 04:42:11 pm »

thanks for helping and explaining it.

the answer to it is : 1. 2/3 2. -1,2,3,6 ( as said in the solutions )

TrueTears · « **Reply #5 on:** March 10, 2010, 04:45:18 pm »

Quote from: m@tty on March 10, 2010, 04:38:48 pm

Quote from: TrueTears on March 10, 2010, 04:33:11 pm
2. p|p-5|=6

from here we see p>0 so it is either 6 or 2,3
Are you being serious? I assume you're not...

$\begin{align}|p^2-5p|&\ne p|p-5| \\ &=|p||p-5| \end{align}$

-1 is also a solution.

WHAT A PRO as u can see im shit at maths i shud stick with accounting

the.watchman · « **Reply #6 on:** March 10, 2010, 05:15:15 pm »

Quote from: TrueTears on March 10, 2010, 04:45:18 pm

Quote from: m@tty on March 10, 2010, 04:38:48 pm
Quote from: TrueTears on March 10, 2010, 04:33:11 pm
2. p|p-5|=6

from here we see p>0 so it is either 6 or 2,3
Are you being serious? I assume you're not...

$\begin{align}|p^2-5p|&\ne p|p-5| \\ &=|p||p-5| \end{align}$

-1 is also a solution.
WHAT A PRO as u can see im shit at maths i shud stick with accounting

LOL asif

(3) (a) $y=\frac{4}{(x-2)^2}-1$ ,

For y-int: Let $x=0$ , $y=\frac{4}{(0-2)^2}-1 = \frac{4}{4}-1 = 0$

For x-int: Let $y=0$ , $0=\frac{4}{(x-2)^2}-1$

$1=\frac{4}{(x-2)^2}$

$(x-2)^2=4$

$(x-2)^2-4=0$

$(x-2+2)(x-2-2)=0$ , $x=0,4$

(b) For $y=\frac{4}{(x-2)^2}-1$

Horizontal asmyptote: $y=-1$

Vertical asymptote: $x-2=0$ , $x=2$

huss48 · « **Reply #7 on:** March 10, 2010, 07:37:17 pm »

thanks for the help Cheers!!

jiaptial · « **Reply #8 on:** March 10, 2010, 07:57:46 pm »

Hey guys, just wondering if someone could help me with this question - I've forgotton logs!

If 2+log₁₀(4x)=log₁₀(y), find y in terms of x.

Thanks

Stroodle · « **Reply #9 on:** March 10, 2010, 08:02:48 pm »

jiaptial · « **Reply #10 on:** March 10, 2010, 08:05:39 pm »

thanks heaps

Stroodle · « **Reply #11 on:** March 10, 2010, 08:07:28 pm »

no probs

moekamo · « **Reply #12 on:** March 11, 2010, 05:18:02 am »

Quote from: jiaptial on March 10, 2010, 07:57:46 pm

Hey guys, just wondering if someone could help me with this question - I've forgotton logs!

If 2+log₁₀(4x)=log₁₀(y), find y in terms of x.

Thanks

another way:

$2+\log_{10} 4x = \log_{10} y$
$\implies 10^{2+\log_{10} 4x} = 10^{ \log_{10} y}$
$\implies 10^{2} \cdot 10^{\log_{10} 4x} = y$
$\implies y= 100 \cdot 4x = 400x \ \forall \ x \in \mathbb{R^{+}}$

note it is only valid for positive $x$ since the initial equation must also be satisfied, and you cant log a negative number!

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Author Topic: need help with these questions (Read 1166 times) Tweet Share

huss48

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