Slope of 9 passes through the point (-3,2)

Mathematics

1 answer:

boyakko [2]2 years ago

7 0

Answer:

y-2=9(x+3)

Step-by-step explanation:

y-y1=m(x-x1)

y-2=9(x-(-3))

y-2=9(x+3)

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Solve 6x^2 - 42 = 0 for the exact values of x

GREYUIT [131]

Answer:

The values of x are

$x=-\sqrt{7} , x=\sqrt{7}$

Step-by-step explanation:

we have

$6x^{2}-42=0$

Solve for x

Adds 42 both sides

$6x^{2}-42+42=0+42$

$6x^{2}=42$

Divide by 6 both sides

$6x^{2}/6=42/6$

simplify

$x^2=7$

square root both sides

$x=\pm\sqrt{7}$

therefore

The values of x are

$x=-\sqrt{7} , x=\sqrt{7}$

6 0

3 years ago

5 Exam-style ABCD is a kite.

Mnenie [13.5K]

let's recall that in a Kite the diagonals meet each other at 90° angles, Check the picture below, so we're looking for the equation of a line that's perpendicular to BD and that passes through (-1 , 3).

keeping in mind that perpendicular lines have negative reciprocal slopes, let's check for the slope of BD

$y = \stackrel{\stackrel{m}{\downarrow }}{3}x-1\impliedby \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}$

$\stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{3\implies \cfrac{3}{1}} ~\hfill \stackrel{reciprocal}{\cfrac{1}{3}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{1}{3}}}$

so we're really looking for the equation of a line whose slope is -1/3 and passes through point A

$(\stackrel{x_1}{-1}~,~\stackrel{y_1}{3})\qquad \qquad \stackrel{slope}{m}\implies -\cfrac{1}{3} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{3}=\stackrel{m}{-\cfrac{1}{3}}[x-\stackrel{x_1}{(-1)}]\implies y-3=-\cfrac{1}{3}(x+1) \\\\\\ y-3=-\cfrac{1}{3}x-\cfrac{1}{3}\implies y=-\cfrac{1}{3}x-\cfrac{1}{3}+3\implies y=-\cfrac{1}{3}x+\cfrac{8}{3}$