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2 years ago

Calculate the slope of each line

Mathematics

1 answer:

quester [9]2 years ago

3 0

Answer:

blue line: 20

black line: -20

Step-by-step explanation:

use rise (y) over run (x)! as the blue line goes from points (8,80) to (9,100) the difference in the x or run is 1 while the difference in the y or rise is 20 which makes it 20/1 = 20. for the black line, because it's going down it indicates it will be a negative slope, use the same trick as last time, our first point is (5,20) and our next point is (6,0) the difference in x is 1 and the difference in y is -20, so -20/1 = -20! hope this helped :-)

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P=?,r=5%,t=13 months, i=$103.55

V125BC [204]

Answer:

$1917.60

Step-by-step explanation:

Interest = Principal x Rate x Time

103.55 = (.05)(13/12)P

103.55 = .054P

P = 1917.60

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3 years ago

Louise skip counts by 4 on a number to.find 5 x 4 how many jumps should she draw in the number line

mixer [17]

I think its 20 but I'm not sure

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3 years ago

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Soloha48 [4]

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2 years ago

Which line(s) have a positive slope ?

mezya [45]

B & c
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6 0

2 years ago

Evaluate the integral following

alina1380 [7]

Answer:

$\displaystyle{4\tan x + \sin 2x - 6x + C}$

Step-by-step explanation:

We are given the integral of:

$\displaystyle{\int 4(\sec x - \cos x)^2 \, dx}$

First, we can use a property to separate a constant out of integrand:

$\displaystyle{4 \int (\sec x - \cos x)^2 \, dx}$

Next, expand the expression (integrand):

$\displaystyle{4 \int \sec^2 x - 2\sec x \cos x + \cos^2 x \, dx}$

Since $\displaystyle{\sec x = \dfrac{1}{\cos x}}$ then it can be simplified to:

$\displaystyle{4 \int \dfrac{1}{\cos^2 x} - 2\dfrac{1}{\cos x} \cos x + \cos^2 x \, dx}\\\\\displaystyle{4 \int \dfrac{1}{\cos^2 x} - 2 + \cos^2 x \, dx}$

Recall the formula:

$\displaystyle{\int \dfrac{1}{\cos ^2 x} \, dx = \int \sec ^2 x \, dx = \tan x + C}\\\\\displaystyle{\int A \, dx = Ax + C \ \ \tt{(A \ and \ C \ are \ constant.)}$

For $\displaystyle{\cos ^2 x}$ , we need to convert to another identity since the integrand does not have a default or specific integration formula. We know that:

$\displaystyle{2\cos^2 x -1 = \cos2x}$

We can solve for $\displaystyle{\cos ^2x}$ which is:

$\displaystyle{2\cos^2 x = \cos2x+1}\\\\\displaystyle{\cos^2x = \dfrac{\cos 2x +1}{2}}$

Therefore, we can write new integral as:

$\displaystyle{4 \int \dfrac{1}{\cos^2 x} - 2 + \dfrac{\cos2x +1}{2} \, dx}$

Evaluate each integral, applying the integration formula:

$\displaystyle{\int \dfrac{1}{\cos^2x} \, dx = \boxed{\tan x + C}}\\\\\displaystyle{\int -2 \, dx = \boxed{-2x + C}}\\\\\displaystyle{\int \dfrac{\cos 2x +1}{2} \, dx = \dfrac{1}{2}\int \cos 2x +1 \, dx}\\\\\displaystyle{= \dfrac{1}{2}\left(\dfrac{1}{2}\sin 2x + x\right) + C}\\\\\displaystyle{= \boxed{\dfrac{1}{4}\sin 2x + \dfrac{1}{2}x + C}}$

Then add all these boxed integrated together then we'll get:

$\displaystyle{4\left(\tan x - 2x + \dfrac{1}{4}\sin 2x + \dfrac{1}{2} x\right) + C}$

Expand 4 in the expression:

$\displaystyle{4\tan x - 8x +\sin 2x + 2 x + C}\\\\\displaystyle{4\tan x + \sin 2x - 6x + C}$

Therefore, the answer is:

$\displaystyle{4\tan x + \sin 2x - 6x + C}$

4 0

1 year ago