All categories

3 years ago

-0.59x+0.29=7.2 HELP ME PLEASE!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!1

Mathematics

2 answers:

IgorLugansk [536]3 years ago

7 0

Answer:

x=-11.7

Step-by-step explanation:

-0.59x+0.29=7.2

subtract 0.29 on both sides to isolate x

-0.59x=6.91

divide -0.59 from both sides (will be a large number)

-11.711864

x=-11.7

kaheart [24]3 years ago

6 0

Answer:

Step-by-step explanation:

-0.59x+0.29=7.2

-0.59x+0.29-0.29=7.2-0.29

-0.59x=7.2

divide both sides by -0.59

x=− 11.7118644

that can be rounded to x= -11.7

You might be interested in

A window washer drops a brush from a scaf-

yuradex [85]

9.8 * 1.53 = 14.994 (speed/velocity)
1/2 * 9.8 * 1.53^2 = 1/2 * 9.8 * 2.3409 = 11.47041 (distance)

8 0

4 years ago

Find the area of the region that lies inside the first curve and outside the second curve.

marishachu [46]

Answer:

Step-by-step explanation:

From the given information:

r = 10 cos( θ)

r = 5

We are to find the the area of the region that lies inside the first curve and outside the second curve.

The first thing we need to do is to determine the intersection of the points in these two curves.

To do that :

let equate the two parameters together

So;

10 cos( θ) = 5

cos( θ) = $\dfrac{1}{2}$

$\theta = -\dfrac{\pi}{3}, \ \ \dfrac{\pi}{3}$

Now, the area of the region that lies inside the first curve and outside the second curve can be determined by finding the integral . i.e

$A = \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} (10 \ cos \ \theta)^2 d \theta - \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \ \ 5^2 d \theta$

$A = \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} 100 \ cos^2 \ \theta d \theta - \dfrac{25}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \ \ d \theta$

$A = 50 \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \begin {pmatrix} \dfrac{cos \ 2 \theta +1}{2} \end {pmatrix} \ \ d \theta - \dfrac{25}{2} \begin {bmatrix} \theta \end {bmatrix}^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}}$

$A =\dfrac{ 50}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \begin {pmatrix} {cos \ 2 \theta +1} \end {pmatrix} \ \ d \theta - \dfrac{25}{2} \begin {bmatrix} \dfrac{\pi}{3} - (- \dfrac{\pi}{3} )\end {bmatrix}$

$A =25 \begin {bmatrix} \dfrac{sin2 \theta }{2} + \theta \end {bmatrix}^{\dfrac{\pi}{3}}_{\dfrac{\pi}{3}} \ \ - \dfrac{25}{2} \begin {bmatrix} \dfrac{2 \pi}{3} \end {bmatrix}$

$A =25 \begin {bmatrix} \dfrac{sin (\dfrac{2 \pi}{3} )}{2}+\dfrac{\pi}{3} - \dfrac{ sin (\dfrac{-2\pi}{3}) }{2}-(-\dfrac{\pi}{3}) \end {bmatrix} - \dfrac{25 \pi}{3}$

$A = 25 \begin{bmatrix} \dfrac{\dfrac{\sqrt{3}}{2} }{2} +\dfrac{\pi}{3} + \dfrac{\dfrac{\sqrt{3}}{2} }{2} + \dfrac{\pi}{3} \end {bmatrix}- \dfrac{ 25 \pi}{3}$

$A = 25 \begin{bmatrix} \dfrac{\sqrt{3}}{2 } +\dfrac{2 \pi}{3} \end {bmatrix}- \dfrac{ 25 \pi}{3}$

$A = \dfrac{25 \sqrt{3}}{2 } +\dfrac{25 \pi}{3}$

The diagrammatic expression showing the area of the region that lies inside the first curve and outside the second curve can be seen in the attached file below.

Download docx

7 0

3 years ago

I am so stuck on this. I need your help. <br>

Aleksandr [31]

You're gonna have to draw your own plot; let's fill in the table.

First let's count and sort the numbers;

2, 9, 10, 12, 14, 14, 15, 22, 32, 36, 43

n = 11

Median is the sixth one, in the middle: 14

Min: 2

Max: 43

Q1: For 11 points, the first quarter is between the 2nd and the 3rd: 9.5

Q3: Between 32 and 36: 34

Outlier? 43 maybe, hard to tell.

Mean: (2+9+10+12+14+14+15+22+32+36+43)/11 = 19

Range: from 2 to 43

IQR: from 9.5 to 39.5

The mean doesn't get plotted in a box and whiskers plot. The left whisker goes to the minimum, the left box Q1, the line through the box the median, the right box edge Q3, and the right whisker the maximum.

I'm not really sure what they're talking about here, how about b for big blue squares, r for blue rectangles and s for small squares:

(5b + 6r + 10s) - (3b + 2r + 7s) = 2b + 4r + 3s

6 0

3 years ago

What is the fraction

liberstina [14]

Answer:

what fraction ??

Step-by-step explanation:

I don't see the problem

4 0

3 years ago

Read 2 more answers

How do I solve this?

laila [671]

The < is just like and equal sign just looks a little different think of it that was for the first couple steps. So you have 13-3y<-5. First do, -5+13=8. Now you have -3y<8. Then your want to do 8 divided by -3 and you get=-2.666666667 and that equals y.

-2.7=y then make that equal sign into the normal sign again. -2.7

7 0

3 years ago

Read 2 more answers