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3 years ago

FAST PLEASE !!!!

Mathematics

1 answer:

zhannawk [14.2K]3 years ago

5 0

Answer:

2/3

Step-by-step explanation:

To get the slope, you do (y2-y1)/(x2-x1).

You can also remember it as rise/run, as it is the change in y/change in x.

You can make either coordinate 1 or 2, just keep it consistent with the rest of the problem.

For this, I chose (-5, -7) to be 2, and (4, -1) to be 1.

Following the formula for slope, (y2-y1)/(x2-x1), you put the values in: (-7-{-1})/(-5-4).

-7 minus -1 is -6 and -5 minus 4 is -9.

Thus you get -6/-9. Don’t forget to simplify! A negaive divided by a negative is positive, and both 6 and 9 can be divided by 3. 3 goes into 6 TWO times and into 9 THREE times.

So, you get a positive fraction, 2/3.

Hope this helped.

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Which expression is equal to 21,340?

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3 years ago

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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

Angela had $6w. AFter buying some books at $17 each, she had $w left.

kkurt [141]

Answer:a) W=-17x/-5

Step-by-step explanation:

Number of books = x

6W-(X×17)=W

6W -17X = W

Collect like terms

-17X= W-6W

-17X = -5W

Divide both sides by the coefficient of W

-17X÷ -5 = -5W÷ -5

W = -17X÷ -5

B) W = 34

W = -17X ÷ -5

34 = -17X ÷ -5

Cross multiply

-17X = 34× -5

-17X = -170

Divide both sides by the coefficient of X

X = -170 ÷ -17

X = 10

∴ Angela bought 10 books

3 0

2 years ago