Write an equation of the line that passes through (3,2), and is perpendicular to the line

Mathematics

1 answer:

frutty [35]3 years ago

8 0

Step-by-step explanation:

step 1. perpendicular lines have negative reciprocal slopes

step 2. perpendicular to m = 1/3 is the slope m = -3/1 or m = -3

step 3. a line through (3, 2) with m = -3 is y - 2 = -3(x - 3)

step 4. y - 2 = -3x + 9

step 5. y = -3x + 11.

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$\large\begin{array}{l} \left\{\!\begin{array}{l} \mathsf{f(x)=x^2-6x+2}\\ \mathsf{g(x)=\sqrt{x}}\\ \end{array}\right. \end{array}$

$\large\begin{array}{l} \textsf{a) }\mathsf{(f\circ g)(x)}\\\\ =\mathsf{f\big[g(x)\big]}\\\\ =\mathsf{\big[g(x)\big]^2-6\cdot g(x)+2}\\\\ =\mathsf{\big[\sqrt{x}\big]^2-6\sqrt{x}+2}\\\\\\ \therefore~~\boxed{\begin{array}{c}\mathsf{(f\circ g)(x)=x-6\sqrt{x}+2} \end{array}}\qquad\checkmark \end{array}$

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$\large\begin{array}{l} \textsf{b) }\mathsf{(g\circ f)(-2)}\\\\ =\mathsf{g\big[f(-2)\big]}\\\\ =\mathsf{\sqrt{f(-2)}}\\\\ =\mathsf{\sqrt{(-2)^2-6\cdot (-2)+2}}\\\\ =\mathsf{\sqrt{4+12+2}}\\\\ =\mathsf{\sqrt{18}}\\\\ =\mathsf{\sqrt{3^2\cdot 2}}\\\\ =\mathsf{3\sqrt{2}}\\\\\\ \therefore~~\boxed{\begin{array}{c}\mathsf{(g\circ f)(-2)=3\sqrt{2}} \end{array}}\qquad\checkmark \end{array}$

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$\large\textsf{I hope this helps. :-)}$

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