Looking at the point on the blue line for f(x) (0,1) and the point on the red line for g(x) (0,-9), the difference of the Y-coordinates is 10 ( from 1 to -9 is 10).

SO to have the red line move down 10 units, you would need to subtract 10 from the original equation.

The correct equation would be the last one g(x) = 5× - 10

4 0

4 years ago

I need this answer ASAP

NikAS [45]

Answer:

D. 7x - 6 = x + 6(x + 5)

Step-by-step explanation:

a equation will have no solution if the final value is not real or not equal

D) 7x - 6 = x + 6(x + 5)

7x-6 = x+6x+30

7x-x-6x-6-30= 0

-36≠0

so -36 ≠ 0 ... so no solution

A. 9x + 9 = 11x – 2x + 9

9x-11x+2x+9-9=0

0=0

have solution, Real, true

B. 7x + 30 = x + 6(x + 5)

7x+30 = x+6x+30

7x-x-6x+30-30= 0

0=0

have solution, Real, true

C. -5x + 9 = 2x + 9 - 7x

-5x-2x+7x+9-9= 0

-7x+7x=0

0=0

have solution, Real, true

7 0

3 years ago

What is the slope-intercept form of linear equations?

lara31 [8.8K]

to get the equation of any straight line, we simply need two points off of it, let's use the points from the picture below.

$(\stackrel{x_1}{8}~,~\stackrel{y_1}{3})\qquad (\stackrel{x_2}{12}~,~\stackrel{y_2}{5}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{5}-\stackrel{y1}{3}}}{\underset{run} {\underset{x_2}{12}-\underset{x_1}{8}}}\implies \cfrac{2}{4}\implies \cfrac{1}{2}$

$\begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{3}=\stackrel{m}{\cfrac{1}{2}}(x-\stackrel{x_1}{8}) \\\\\\ y-3=\cfrac{1}{2}x-4\implies y=\cfrac{1}{2}x-1$

if we already have the slope, and we can see the y-intercept on the table, then we can simply use the slopel-intercept form and plug both of them in.

$\begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}\qquad \begin{array}{|c|ll} \cline{1-1} slope\\ \cline{1-1} \cfrac{1}{2}\\ \cline{1-1} y-intercept&\\\cline{1-1} (0~~,~~-1)\\ \cline{1-1} \end{array}~\hfill y=\cfrac{1}{2}x-1$

6 0

2 years ago

Help please!!!! if <1 <3 which conclusion can be made?

netineya [11]

Answer:

B. c // d

Step-by-step explanation:

When ∠1 = ∠3, it means that they are corresponding angles and for that to be true, there must be a pair of parallel lines and one line to cut across both of the lines. Thus, we can conclude that c and d are the parallel lines and a is the line that cut across both of them, creating the corresponding angles.

8 0

3 years ago