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2 years ago

You move up 5 units and down 5 units. You end at (4, -3). Where did you start?

Mathematics

2 answers:

aliina [53]2 years ago

7 0

Answer:

(4, -3) duh

Step-by-step explanation:

Margaret [11]2 years ago

4 0

Answer: 1-,2

Step-by-step explanation: Work backwards.

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Nasir has an annual salary of 64,000 and his company pays him twice a month. What is the gross income per paycheck that Nasir re

ryzh [129]

64,000/2=32,000

32,000/12= $2,666.67 per paycheck

5 0

3 years ago

What is the equation of this trend line?<br> RE<br> Enter your answers b y filling in the boxes.

Deffense [45]

Answer:

$K=-2J+28$

Step-by-step explanation:

step 1

Find the slope of the line

The formula to calculate the slope between two points is equal to

$m=\frac{k2-k1}{j2-j1}$

we have the points

(2,24) and (6,16)

substitute in the formula

$m=\frac{16-24}{6-2}$

$m=\frac{-8}{4}$

$m-2$

step 2

Find the equation of the line in slope intercept form

$K=mJ+b$

we have

$m=-2$

$b=28$ ----> the k-intercept is the point (0,28) see the graph

substitute

$K=-2J+28$

7 0

3 years ago

6p+7=-9 how to solve for p.

lions [1.4K]

Step-by-step explanation:

6p+7=9

6p=9-7

6p=2

p=2/6

p=1/2

8 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:

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