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3 years ago

(7,-4) and (2,-4) find the slope between each pair of points

Mathematics

2 answers:

liq [111]3 years ago

8 0

M = 0
There is no slope in this problem

MatroZZZ [7]3 years ago

6 0

Answer:

the slope is 0

Step-by-step explanation:

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Among two supplementary angles, the measure of the larger angle is 44°

zhuklara [117]

Answer:

68° and 112°

Step-by-step explanation:

Supplementary angles sum to 180°

let the smaller angle be x then the larger angle is x + 44 , then

x + x + 44 = 180 , that is

2x + 44 = 180 ( subtract 44 from both sides )

2x = 136 ( divide both sides by 2 )

x = 68

Smaller angle = 68° and

larger angle = x + 44 = 68 + 44 = 112°

6 0

3 years ago

‼️‼️‼️‼️‼️‼️‼️‼️‼️‼️‼️Please help, I have been waiting for over an hour now and still no one has helped.

Aliun [14]

It would be 76% because if you were to put the shaded box in the bottom corner there would be way more than 50% not covered hope this helped

5 0

4 years ago

Read 2 more answers

There are 32 students in Joe's class. Of the 32 students, 8 students ride their bikes to school, 5 walk to school, 4 ride to sch

vampirchik [111]

Answer:

5/32 (aka option A) should be the correct answer!

Step-by-step explanation:

ok so... its 5/32 who walks to school!

since there are 5 students who walked to school, and 32 students are in the class, 5/32 should be the correct answer!

3 0

3 years ago

Read 2 more answers

Find the slope of the line that contains the points:<br> (-4,3) and (4,9)

Stolb23 [73]

I believe the slope is 3/4

9-3 = 6
4 - (-4) = 8

Simplify to get 3/4

3 0

3 years ago

Read 2 more answers

What is the solution of

kobusy [5.1K]

Answer:

Third option: x=0 and x=16

Step-by-step explanation:

$\sqrt{2x+4}-\sqrt{x}=2$

Isolating √(2x+4): Addind √x both sides of the equation:

$\sqrt{2x+4}-\sqrt{x}+\sqrt{x}=2+\sqrt{x}\\ \sqrt{2x+4}=2+\sqrt{x}$

Squaring both sides of the equation:

$(\sqrt{2x+4})^{2}=(2+\sqrt{x})^{2}$

Simplifying on the left side, and applying on the right side the formula:

$(a+b)^{2}=a^{2}+2ab+b^{2}; a=2, b=\sqrt{x}$

$2x+4=(2)^{2}+2(2)(\sqrt{x})+(\sqrt{x})^{2}\\ 2x+4=4+4\sqrt{x}+x$

Isolating the term with √x on the right side of the equation: Subtracting 4 and x from both sides of the equation:

$2x+4-4-x=4+4\sqrt{x}+x-4-x\\ x=4\sqrt{x}$

Squaring both sides of the equation:

$(x)^{2}=(4\sqrt{x})^{2}\\ x^{2}=(4)^{2}(\sqrt{x})^{2}\\ x^{2}=16 x$

This is a quadratic equation. Equaling to zero: Subtract 16x from both sides of the equation:

$x^{2}-16x=16x-16x\\ x^{2}-16x=0$

Factoring: Common factor x:

x (x-16)=0

Two solutions:

1) x=0

2) x-16=0

Solving for x: Adding 16 both sides of the equation:

x-16+16=0+16

x=16

Let's prove the solutions in the orignal equation:

1) x=0:

$\sqrt{2x+4}-\sqrt{x}=2\\ \sqrt{2(0)+4}-\sqrt{0}=2\\ \sqrt{0+4}-0=2\\ \sqrt{4}=2\\ 2=2$

x=0 is a solution

2) x=16

$\sqrt{2x+4}-\sqrt{x}=2\\ \sqrt{2(16)+4}-\sqrt{16}=2\\ \sqrt{32+4}-4=2\\ \sqrt{36}-4=2\\ 6-4=2\\ 2=2$

x=16 is a solution

Then the solutions are x=0 and x=16

5 0

3 years ago