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3 years ago

Solve for q. k = 4pq²

Mathematics

2 answers:

9966 [12]3 years ago

7 0

K = 4pq^2
k/4p = q^2
0.5(k/p)^1/2 = q

Sphinxa [80]3 years ago

4 0

Divide both sides by 4p, then take the square root on both sides, then get the root out of the denominator on the left side. You should get $\dfrac{\sqrt{k}\times\sqrt{p}}{2}=q$

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What is the equation of the line that is parallel to the line 2x+3y=-8 and passes through the point (2,-2)?

Leya [2.2K]

Equation of line passing through (2, -2) and parallel to 2x+3y = -8 is $y=\frac{-2 x}{3}+\frac{-2}{3}$

<h3><u>Solution:</u></h3>

Need to write equation of line parallel to 2x+3y=-8 and passes through the point (2, -2)

Generic slope intercept form of a line is given by y = mx + c

where "m" = slope of the line and "c" is the y - intercept

Let’s first find slope intercept form of 2x+3y=-8 to get slope of line

$\begin{array}{l}{2 x+3 y=-8} \\\\ {=>y=\frac{-2 x-8}{3}} \\\\ {\Rightarrow y=-\frac{2}{3} x-\frac{8}{3}}\end{array}$

On comparing above slope intercept form of given equation with generic slope intercept form y = mx + c,

$\text {for line } 2 x+3 y=-8, \text { slope } m=-\frac{2}{3}$

We know that slopes of parallel lines are always equal

So the slope of line passing through (2, -2) is also $m=-\frac{2}{3}$

Equation of line passing through $(x_1 , y_1)$ and having slope of m is given by

$\left(y-y_{1}\right)=\mathrm{m}\left(x-x_{1}\right)$

$\text { In our case } x_{1}=2 \text { and } y_{1}=-2$

Substituting the values in equation of line we get

$(y-(-2))=-\frac{2}{3}(x-2)$

$\begin{array}{l}{\Rightarrow y+2=\frac{-2 x+4}{3}} \\\\ {=>3(y+2)=-2 x+4} \\\\ {=>3 y+6=-2 x+4} \\\\ {3 y=-2 x-2}\end{array}$

$y=\frac{-2 x}{3}+\frac{-2}{3}$

Hence equation of line passing through (2 , -2) and parallel to 2x + 3y = -8 is given as $y=\frac{-2 x}{3}+\frac{-2}{3}$

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