$f(\stackrel{x_1}{-4})=\stackrel{y_1}{0}\qquad f(\stackrel{x_2}{0})=\stackrel{y_2}{0}~\hfill (\stackrel{x_1}{-4}~,~\stackrel{y_1}{0})\qquad (\stackrel{x_2}{0}~,~\stackrel{y_2}{0}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{0}-\stackrel{y1}{0}}}{\underset{run} {\underset{x_2}{0}-\underset{x_1}{(-4)}}}\implies \cfrac{0}{0+4}\implies \cfrac{0}{4}\implies 0$

$\begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{0}=\stackrel{m}{0}(x-\stackrel{x_1}{(-4)})\implies \boxed{y = 0}$

6 0

3 years ago

Which of the following transformations would prove that the corresponding angles 1. a translation of line AI by the vector BF<br

Sedaia [141]

Answer:

elemenope

Step-by-step explanation:

6 0

3 years ago

Factor this polynomial: x^2 + 6x + 9<br> (x+3)(x+3)<br> (x-3) (x-3)<br> (x-3) (x+3)<br> (x+3) (x-3)

Lyrx [107]

Answer:

(x+3)(x+3)

Step-by-step explanation:

1) this polynomial is in the form of ax^2+bx+c

c=9

b=6

a=1

what*what=c(9)

and what+what=b(6)

3 is the answer to both questions:

now substitute "3x" in place of "6"

x^2+3x+3x+9

split in 2 groups:

group a) (x^2+3x)

gcf of group a=x

x(x+3)

group b) (3x+9)

gcf of group b=3

3(x+3)

combine the numbers OUTSIDE the parantheses

(x+3)(x+3)

6 0

3 years ago

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if a number is increased by 5 and the result is then divided by three,the result is seven. write an equation that models this ve

gregori [183]

A number (we don’t know the number so I’ll use X) plus five. So we can start the equation of with X+5 then divided by 3=7 so X+5➗3=7 so to solve what X is we can just go in reverse. 7 TIMES 3= 21-5=16 so X=16

6 0

3 years ago

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