All categories

2 years ago

HELP PLEASE ILL GIVE THE BRANLIEST THING IF U DO, PLEASEE

Mathematics

1 answer:

bezimeni [28]2 years ago

8 0

Answer:

The y-intercept is 5, and the slope is -3/2, y=5

Step-by-step explanation:

You might be interested in

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

What is 79 881/1000 as decimal

den301095 [7]

79.881 hope this helps
i might be wrong

6 0

3 years ago

What is the prime number of 30 to 59

Alinara [238K]

Answer:

the prime numbers between 30 and 59 are 31, 37, 41, 43, 47, 53

6 0

3 years ago

A teacher reviews 4-1/2 papers per hour, how many papers will that teacher review in 6-1/3 hours

MrRa [10]

Answer:

$28\frac{1}{2}\ papers$

Step-by-step explanation:

step 1

Convert mixed numbers to an improper fractions

$4\frac{1}{2}=\frac{4*2+1}{2}=\frac{9}{2}$

$6\frac{1}{3}=\frac{6*3+1}{3}=\frac{19}{3}$

step 2

we know that

A teacher reviews 9/2 papers per hour

using proportion

Find how many papers will the teacher review in 19/3 hours

Let

x-----> the number of papers

$(9/2)/1=x/(19/3)$

$x=(19/3)(9/2)$

$x=28.5\ papers$

Convert to mixed numbers

$28.5=28\frac{1}{2}\ papers$

5 0

2 years ago

Which is the side length of a cube with a volume of 1331 m3?

suter [353]

Let's remember the equation for the volume of a cube:
V= x^3 where x is the length of one side
Since a cube has equal length for length, width, and height.

Now, use what you're given
V= 1331

And put that in terms of x
x^3 = 1331

Now solve for x!
x= cube root (1331)

Think about it! cube root of 1331 * cube root of 1331 * cube root of 1331... will equal 1331 m^3!

Hope this helps

8 0

2 years ago