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3 years ago

Of all the students enrolled at Anderson High School, 9%, percent are in the band. The band has 180 members.

Mathematics

1 answer:

Mashutka [201]3 years ago

7 0

If you're asking about how many are in the school, the answer is 2000

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A 4500 kg vehicle and a 3700kg vehicle are both moving at a speed of 85m/s toward east.Calculate the momentum of each veheicle?

lana [24]

The momentum formula is
p = m*v

where,
p = momentum (not sure why p is chosen but many physics books tend to pick this letter)
m = mass in kilograms
v = velocity in meters per second

So let's calculate the momentum for the 4500 kg vehicle
p = m*v
p = 4500*85
p = 382500 kg m/s

And let's do the same for the 3700 kg vehicle
p = m*v
p = 3700*85
p = 314500 kg m/s

8 0

4 years ago

I agree because they are right I don't really care for points btw

grigory [225]

Answer:

Thanks :)

Step-by-step explanation:

6 0

3 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 need help graphing 2x - 6y = 42 I'm just really lazy

solong [7]

Answer:

We can use slope intercept form to get the points needed. Y= -7+1/3x The points are (0,-7) and (3,-6)

Step-by-step explanation:

Subtract 2x from the left side and place it over to the right side with the 42. Now we have -6y= 42-2x. From here we divide by -6 and we get y= -7+1/3x. We know that are slope is 1/3 which the one is the rise and the 3 is the run. We also know that our y intercept is -7. We plot the points at (0,-7) and (3,-6)

3 0

3 years ago

Expand $\left(x^2 \frac{1}{x}\right)^3$. (Write the terms with higher degree first, so for example an $x^2$ term would come befo

Ilya [14]

Answer:

$x^6+3x^3+3+\dfrac{1}{x^3}$

Step-by-step explanation:

We can use the pascal triangle to find the coefficients of the expasion of the cube. It holds that

$(x^2 + \dfrac{1}{x})^3=(x^2)^3+3(x^2)^2\dfrac{1}{x}+3x^2(\dfrac{1}{x})^2+(\dfrac{1}{x})^3\\\\=x^6+3x^4\dfrac{1}{x}+3x^2\dfrac{1}{x^2}+\dfrac{1}{x^3}\\\\=x^6+3x^3+3+\dfrac{1}{x^3}$

4 0

4 years ago