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2 years ago

8

What is the slope of this line? A) 1 B) 0 C) No slope D) -2

2 answers:

leva [86]2 years ago

5 0

Answer:

0

Step-by-step explanation:

VladimirAG [237]2 years ago

5 0

Hey there! I'm happy to help!

The slope is the incline of a line, or how steep it is.

If a slope is positive, it is going upwards, and if it is negative, it will be going downwards.

We see that our line is going horizontally, so answers A and D are incorrect as it is not going diagonally upwards or downwards.

If a line is vertical, it has no slope. Since it is only going up and down and not side to side, there is no slope. However, this line is horizontal, so it still has a slope. Answer C is incorrect.

This line has a slope of B) 0. This is because our line is going side to side, but it isn't going up. So, our slope is just flat, or a slope of zero.

Have a wonderful day and keep on learning! :D

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kompoz [17]

Answer:

The answer is 5/36

Step-by-step explanation:

first, you are going to convert both of the fractions to have the same dinominator by multiplying by any of their factors.

32/36 - 27/36 = 536

Then you subtract the numerators and that's it

hope I helped :)

7 0

2 years ago

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Lynna [10]

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2 years ago

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statuscvo [17]

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2 years ago

19x+rx= -37x+w what is the x

seraphim [82]

Answer:

x = w / (56 + r)

Step-by-step explanation:

Given the equation :

19x + rx= -37x + w ; find x

Collecting like terms

19x + rx + 37x = w

Factorizing x in the Left hand side

x(19 + 37 + r) = w

x(56 + r) = w

Therefore we can obtain x by dividing both sides by (56 + r)

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5 0

2 years ago

Solve using long division <br> Please

madreJ [45]

1. $Solution,\frac{2x^3+4x^2-5}{x+3}:\quad 2x^2-2x+6-\frac{23}{x+3}$

$Steps:$

$\mathrm{Divide}\:\frac{2x^3+4x^2-5}{x+3}:\quad \frac{2x^3+4x^2-5}{x+3}=2x^2+\frac{-2x^2-5}{x+3}$

$\mathrm{Divide}\:\frac{-2x^2-5}{x+3}:\quad \frac{-2x^2-5}{x+3}=-2x+\frac{6x-5}{x+3}$

$\mathrm{Divide}\:\frac{6x-5}{x+3}:\quad \frac{6x-5}{x+3}=6+\frac{-23}{x+3}$

$\mathrm{Simplify}, =2x^2-2x+6-\frac{23}{x+3}$

$\mathrm{The\:Correct\:Answer\:is\:2x^2-2x+6-\frac{23}{x+3}}$

2. $Solution, \frac{4x^3-2x^2-3}{2x^2-1}:\quad 2x-1+\frac{2x-4}{2x^2-1}$

$Steps:$

$\mathrm{Divide}\:\frac{4x^3-2x^2-3}{2x^2-1}:\quad \frac{4x^3-2x^2-3}{2x^2-1}=2x+\frac{-2x^2+2x-3}{2x^2-1}$

$\mathrm{Divide}\:\frac{-2x^2+2x-3}{2x^2-1}:\quad \frac{-2x^2+2x-3}{2x^2-1}=-1+\frac{2x-4}{2x^2-1}$

$\mathrm{The\:Correct\:Answer\:is\:2x-1+\frac{2x-4}{2x^2-1}}$

$\mathrm{Hope\:This\:Helps!!!}$

$\mathrm{-Austint1414}$

4 0

2 years ago