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3 years ago

What is the slope of the line that passes through the points (-8, 8) and (8, –5)?

Mathematics

1 answer:

True [87]3 years ago

6 0

Answer:

-13/16

Step-by-step explanation:

slope = rise/run = (-5 - 8)/(8 - (-8)) = -13/16

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Please help meeee I don't know this

Vanyuwa [196]

Answer:

$\boxed {m = -\frac{5}{7}}$

Step-by-step explanation:

Use the Slope Formula to determine the slope of two given points:

$m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$

First Point: $(x_{1} , y_{1})$

Second Point: $(x_{2}, y_{2})$

-Substitute both points:

First Point: $(5, 11)$

Second Point: $(19, -1)$

$m = \frac{-1 - 11}{19 - 5}$

-Solve for the slope:

$m = \frac{-1 - 11}{19 - 5}$

$m = \frac{-10}{14}$

$\boxed {m = -\frac{5}{7}}$

Therefore, the slope is $-\frac{5}{7}$

4 0

3 years ago

HELP ASAP!!! (The problem is on a screenshot)

Darina [25.2K]

The correct answers are A, D, and E.

7 0

3 years ago

What are the approximate values of the non-integral roots of the polynomial equation?

irga5000 [103]

Answer:

-4, -2, -3-2i, -3+2i

Step-by-step explanation:

The of the polynomial equation are those values of x, for which f(x)=0.

Consider equation

$f(x)=0\\ \\(x^2+6x+8)(x^2+6x+13)=0$

By zero product property,

$x^2+6x+8=0\ \text{or}\ x^2+6x+13=0$

Solve each equation:

1. $x^2+6x+8=0$

$D=6^2-4\cdot 8=36-32=4\\ \\x_{1,2}=\dfrac{-6\pm \sqrt{4}}{2}=\dfrac{-6\pm 2}{2}=-4,\ -2$

2. $x^2+6x+13=0$

$D=6^2-4\cdot 13=36-52=-16=16i^2\\ \\x_{1,2}=\dfrac{-6\pm \sqrt{16i^2}}{2}=\dfrac{-6\pm 4i}{2}=-3-2i,\ -3+2i$

5 0

4 years ago

Read 2 more answers

--- What is the slope of the equation y=5/4 x -7/4

taurus [48]

Answer:

5/4

Step-by-step explanation:

The slope intercept form for a linear equation is y = mx + b. The variable's (x) coefficient, which is represented as m, is the slope of any linear equation.

5 0

3 years ago

1) 1/3 2)6 3)-6 4)-3 5)-1/3 6) 3

slega [8]

Answer:

$\frac{3}{1}$ or 3

Step-by-step explanation:

To find the slope find two coordinates and set up the equation $\frac{y_{2}-y_{1} }{x_{2} -x_{1} }$

For the coordinates (0,2) and (2,8) y2 is 8 and y1 is 2. x2 is 2 and x1 is 0. $\frac{8_{}-2_ }{2-0_ }$ you have to do subtraction first. 8-2 = 6 and 2-0 = 2

Now we have the equation $\frac{6}{2}$

this can be simplified to $\frac{3}{1}$

6 0

3 years ago