$\bf \displaystyle\int\limits_{-2}^{2}~[\stackrel{\textit{above function}}{(10-x^2)}~~-~~\stackrel{\textit{below function}}{(x^2+2)}]dx\implies \int\limits_{-2}^{2}~(10-x^2-x^2-2)dx \\\\\\ \displaystyle\int\limits_{-2}^{2}~(8-2x^2)dx\implies \left. 8x\cfrac{}{} \right]_{-2}^{2}-\left. \cfrac{2x^3}{3} \right]_{-2}^{2} \\\\\\ ([16]-[-16])~~-~~\left( \left[ \cfrac{16}{3} \right]-\left[ -\cfrac{16}{3} \right]\right)\implies (32)~~-~~\left( \cfrac{32}{3} \right) \\\\\\ \cfrac{64}{3}\implies 21\frac{1}{3}$

5 0

3 years ago

Write an equation that goes through (8,1) and is perpendicular to 2y + 4x =12

Serggg [28]

Answer:

2y -x + 6 = 0

Step-by-step explanation:

Here a equation and a point is given to us and we are interested in finding a equation which is perpendicular to the given equation and passes through the given point .

The given equation is ,

$\sf \longrightarrow 2y + 4x = 12 \\$

$\sf \longrightarrow 4x + 2y = 12$

Firstly convert this into slope intercept form , to find out the slope of the line.

$\sf \longrightarrow 2y = -4x +12\\$

$\sf \longrightarrow y =\dfrac{-4x+12}{2}\\$

$\sf \longrightarrow y =\dfrac{-4x}{2}+\dfrac{12}{2}\\$

$\sf \longrightarrow \red{ y = -2x + 6 }$

Now on comparing it to slope intercept form which is y = mx + c , we have ,

m = -2

And as we know that the product of slopes of two perpendicular lines is -1 . So the slope of the perpendicular line will be negative reciprocal of the slope of the given line. Therefore ,

$\sf \longrightarrow m_{\perp}=\dfrac{-1}{-2}=\red{\dfrac{1}{2}}$