Need Some Help with Inverse and Direct Variation

**bahamamath** · September 5th 2012, 03:32 PM

Hey guys haven’t been here in a while but have a few questions , I realize you all giving me the answers plainly wouldn’t really help me but if you could take me step by step into solving them if you can that would be help

1. Y varies directly as x and y is 2 when x is 3 calculate

A) Y when X is 9

B) X when y = 18

Copy and Complete the following table so that y is directly proportionate to x squared

x - 0 , BLANK , 3 , 4 , 5 , Blank

y- Blank , 12 , 27, blank , 75 , 192

Skip to some of the harder ones...

p is inversely proportionate to the square root of y . If p = 1.2 when y = 100, calculate
A) The value of p when y = 4
B) the value of y when p = 3

(27/8)^-2/3
Solve the following questions

3 ^2x= 243

Any assistance you give is quite esteemed… Thanks all

**skeeter** · September 5th 2012, 05:06 PM

Originally Posted by bahamamath

Hey guys haven’t been here in a while but have a few questions , I realize you all giving me the answers plainly wouldn’t really help me but if you could take me step by step into solving them if you can that would be help

1. Y varies directly as x and y is 2 when x is 3 calculate

y = kx ... sub in your given values for x and y , then solve for k, the constant of proportionality. use your equation to answer A) and B)

A) Y when X is 9

B) X when y = 18

Copy and Complete the following table so that y is directly proportionate to x squared

y = kx² ... same drill as before

x - 0 , BLANK , 3 , 4 , 5 , Blank

y- Blank , 12 , 27, blank , 75 , 192

Skip to some of the harder ones...

p is inversely proportionate to the square root of y . If p = 1.2 when y = 100, calculate

$\color{red}{p = \frac{k}{\sqrt{y}}}$ ... same drill again

A) The value of p when y = 4
B) the value of y when p = 3

(27/8)^-2/3 = (8/27)^2/3 ... cube root of 8/27, then square the result

Solve the following questions

3 ^2x= 243

note that 243 = 3⁵

...

**Soroban** · September 5th 2012, 05:09 PM

Hello, bahamamath!

Here are the first two . . .

$1.\;y\text{ varies directly as }x,\text{ and }y=2\text{ when }x = 3.$

$\text{Calculate:}$

$A)\;y\text{ when }x = 9.$

$y\text{ varies directly as }x\quad\Rightarrow\quad y \:=\:kx$

$y = 2\text{ when }x = 3 \quad\Rightarrow\quad 2 \:=\:k(3) \qyad\Rightarrow\quad k \:=\:\tfrac{2}{3}$

. . $\text{Hence: }\:y \:=\:\tfrac{2}{3}x$

$\text{When }x = 9,\text{ we have: }\:y \:=\:\tfrac{2}{3}(9) \quad\Rightarrow\quad y \:=\:6$

$B)\;x\text{ when }y = 18.$

$\text{When }y = 18,\text{ we have: }\:18 \:=\:\tfrac{2}{3}x \quad\Rightarrow\quad x \:=\:27$

$\text{Copy and complete the following table so that }y\text{ is directly proportionate to }x^2.$

. . $\begin{array}{|c||c|c|c|c|c|c|} \hline x & 0 & - & 3 & 4 & 5 & - \\ \hline y & - &12 & 27& - & 75 & 192 \\ \hline \end{array}$

$y\text{ is directly proportionate to }x^2 \quad\Rightarrow\quad y \:=\:kx^2$

$\text{When }x = 3,\:y = 27\!:\;\;27 \:=\:k(3^2) \quad\Rightarrow\quad k \:=\:3$

. . $\text{Hence: }\:y \:=\:3x^2$

$\text{When }x=0\!:\;y \,=\,3(0^2) \quad\Rightarrow\quad y \,=\,0$

$\text{When }y = 12\!:\;12 \,=\,3x^2 \quad\Rightarrow\quad x^2 \,=\,4 \quad\Rightarrow\quad x \,=\,2$

$\text{When }x = 4\!:\;y \,=\,3(4^2) \quad\Rightarrow\quad y \,=\,48$

$\text{When }y = 192\!:\;192 \,=\,3x^2 \quad\Rightarrow\quad x^2 \,=\,64 \quad\Rightarrow\quad x \,=\,8$

The completed table:

. . $\begin{array}{|c||c|c|c|c|c|c|}\hline x & 0 & 2 & 3 & 4 & 5 & 8 \\ \hline y & 0 &12 & 27& 48 & 75 & 192 \\ \hline\end{array}$

Too slow . . . again!
.

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