$a.\ m=2,\ \text{the line is rising}\\\\b.\ m=-\dfrac{5}{4},\ \text{the line is falling}\\\\c.\ m=0,\ \text{the line is horizontal}\\\\d.\ m\ is\ unde fined,\ \text{the line is vertical}$

Step-by-step explanation:

The formula of a slope:

$m=\dfrac{y_2-y_1}{x_2-x_1}$

m > 0, then a line is rising

m < 0, then a line is falling

m = 0, then a line is horizontal

m is undefined, then a line is vertical

(2, 1) and (4, 5)

$m=\dfrac{5-1}{4-2}=\dfrac{4}{2}=2>0\to\text{rising}$

(-1, 0) and (3, -5)

$m=\dfrac{-5-0}{3-(-1)}=\dfrac{-5}{4}=-\dfrac{5}{4}$

(2, 1) and (-3, 1)

$m=\dfrac{1-1}{-3-2}=\dfrac{0}{-5}=0\to\text{horizontal}$

(-1, 2) and (-1, -5)

$m=\dfrac{-5-2}{-1-(-1))}=\dfrac{-7}{0}\ \text{UNDEFINED}\to\text{vertical}$

7 0

3 years ago

Which represents the solution(s) of the graphed system of equations, y = x2 – 2x and y = –2x – 1? (1, –1) (0, 0) and (0, –1) (0,

olganol [36]

ANSWER

No solution

EXPLANATION

The first equation is

$y = {x}^{2} - 2x$

and the second equation is

$y = - 2x - 1$

We equate the two equations to obtain;

${x}^{2} - 2x = - 2x - 1$

This implies that

${x}^{2} = - 2x + 2x - 1$

${x}^{2} = - 1$

There is no real number whose square is -1.

Therefore, the equation has no solution.

6 0

3 years ago

Read 2 more answers

Help plzzz, you could just tell me to go left or right

Lerok [7]

I ussually think of them as arrows telling me where to go.

if it is > you go ----->

if it is < you go <------

6 0

3 years ago

<img src="https://tex.z-dn.net/?f=%5Csqrt%7B8ux%5E%7B5%7D%20%7D%20%5Csqrt%7B2ux%7D" id="TexFormula1" title="\sqrt{8ux^{5} } \sqr

lozanna [386]

Answer:

4ux³

Step-by-step explanation:

√8ux⁵ √2ux

√8*2 u*u x⁵*x

√16u²x⁶

4ux³

3 0

3 years ago