All categories

2 years ago

Find the slope of the line passing through the points (-5,2) and (-8,8).

Mathematics

1 answer:

sattari [20]2 years ago

7 0

Answer:

slope = -2

Step-by-step explanation:

(-5,2) (-8,8)

equation to find slope: $\frac{y1 - y2}{x1 - x2}$

$\frac{2-8}{-5+8}$ = $\frac{-6}{3}$ = $\frac{-2}{1}$ = -2

You might be interested in

HELP QUICK AND FAST WHAT THE ANSWER

pychu [463]

The answer is C.

V= 3.14 * (7*7) * 14 / 3

6 0

3 years ago

What is the domain and range of the relation shown?

Helga [31]

Answer:

{-4 ≤ x ≤ 4}

{-4 ≤ y ≤ 4}

Step-by-step explanation:

We’ll domain is the amount of x values,

Range is the amount of y values

_______________________________

Domain:

Starts from -4 to 4

{-4 ≤ x ≤ 4}

I made the sign less than or equal to because the circle lines are solid.

Range:

This starts from -4 to 4 also.

{-4 ≤ y ≤ 4}

Thus,

answer choices A. is correct

Hope this helps :)

5 0

3 years ago

Read 2 more answers

Solve the equation: x2 - 12x = -36

Natalija [7]

$x^2-12x=-36\ \ \ |add\ 36\ to\ both\ sides\\\\x^2-12x+36=0\\\\x^2-2x\cdot6+6^2=0\ \ |use\ (a-b)^2=a^2-2ab+b^2\\\\(x-6)^2=0\iff x-6=0\ \ \ |add\ 6\ to\ both\ sides\\\\\boxed{x=6}$

6 0

3 years ago

Read 2 more answers

The length of a rectangle is 4 cm longer than the width, The perimeter is 80 cm. Find the length and the width. Substition Metho

MrRa [10]

Answer: Length = l

Width = w

2l + 2w = 80cm

l+w = 40cm

l = w+4

w+4 + w = 40cm

2w + 4 = 40cm

2w = 36cm

w = 18cm

l = 22cm

jd3sp4o0

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

In a sample of 42 water specimens taken from a construction site, 26 contained detectable levels of lead. Construct 90% confiden

hram777 [196]

Answer:

(0.4958, 0.7422)

Step-by-step explanation:

Let p be the true proportion of water specimens that contain detectable levels of lead. The point estimate for p is $\hat{p}=26/42=0.6190$ . The estimated standard deviation is given by $\sqrt{\hat{p}(1-\hat{p})/n}=\sqrt{(0.6190)(1-0.6190)/42}=0.0749$ . Because we have a large sample, the 90% confidence interval for p is given by $0.6190\pm z_{0.05}0.0749$ where $z_{0.05}=1.6448$ is the value that satisfies that above this and under the standard normal density there is an area of 0.05. So, the confidence interval is $0.6190\pm (1.6448)(0.0749)$ , i.e., (0.4958, 0.7422).

5 0

3 years ago