Which is the equation of a line that has a slope of Negative two-thirds and passes through point (–3,

Evaluate the integral:

Bogdan [553]

Substitute x = √7 sin(t) and dx = √7 cos(t) dt. Then

∫ √(7 - x²) dx = ∫ √(7 - (√7 sin(t))²) • √7 cos(t) dt

… = √7 ∫ √(7 - 7 sin²(t)) cos(t) dt

… = 7 ∫ √(1 - sin²(t)) cos(t) dt

… = 7 ∫ √(cos²(t)) cos(t) dt

We require that -π/2 ≤ t ≤ π/2 in order for the substitution we made to be reversible. Over this domain, cos(t) ≥ 0, so

√(cos²(t)) = |cos(t)| = cos(t)

and the integral reduces to

… = 7 ∫ cos²(t) dt

Recall the half-angle identity for cosine:

cos²(t) = (1 + cos(2t))/2

Then the integral is

… = 7/2 ∫ (1 + cos(2t)) dt

… = 7/2 (t + 1/2 sin(2t)) + C

… = 7t/2 + 7/4 sin(2t) + C

Get the antiderivative back in terms of x. Recall the double angle identity for sine:

sin(2t) = 2 sin(t) cos(t)

We have t = arcsin(x/√7), which gives

sin(t) = sin(arcsin(x/√7)) = x/√7

cos(t) = cos(arcsin(x/√7)) = √(7 - x²)/√7

Then

∫ √(7 - x²) dx = 7/2 arcsin(x/√7) + 7/4 • 2 sin(arcsin(x/√7)) cos(arcsin(x/√7)) + C

… = 7/2 arcsin(x/√7) + x/2 √(7 - x²) + C

5 0

2 years ago

-5(2b + 7) + b < -b - 11

Stells [14]

-5(2b+7) + b < -b - 11
-10b - 35 + b < -b - 11

- 10b + b - 35 < -b - 11

-9b - 35 < -b - 11

- 9b + b < - 11 + 35

-8b < 24

 $\frac{-8b}{-8} \ \textless \ \frac{24}{-8}$

b < -3

Have a nice days.........

8 0

3 years ago

CAN SOMEONE HELP ME IN THIS INTEGRAL QUESTION PLS

finlep [7]

Due to the symmetry of the paraboloid about the z-axis, you can treat this is a surface of revolution. Consider the curve $y=x^2$ , with $1\le x\le2$ , and revolve it about the y-axis. The area of the resulting surface is then

$\displaystyle2\pi\int_1^2x\sqrt{1+(y')^2}\,\mathrm dx=2\pi\int_1^2x\sqrt{1+4x^2}\,\mathrm dx=\frac{(17^{3/2}-5^{3/2})\pi}6$

But perhaps you'd like the surface integral treatment. Parameterize the surface by

$\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec\jmath+u^2\,\vec k$

with $1\le u\le2$ and $0\le v\le2\pi$ , where the third component follows from

$z=x^2+y^2=(u\cos v)^2+(u\sin v)^2=u^2$

Take the normal vector to the surface to be

$\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial u}=-2u^2\cos v\,\vec\imath-2u^2\sin v\,\vec\jmath+u\,\vec k$

The precise order of the partial derivatives doesn't matter, because we're ultimately interested in the magnitude of the cross product:

$\left\|\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}\right\|=u\sqrt{1+4u^2}$

Then the area of the surface is

$\displaystyle\int_0^{2\pi}\int_1^2\left\|\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}\right\|\,\mathrm du\,\mathrm dv=\int_0^{2\pi}\int_1^2u\sqrt{1+4u^2}\,\mathrm du\,\mathrm dv$

which reduces to the integral used in the surface-of-revolution setup.

7 0

3 years ago

I will give brainliest if you solve #11 please and thank you!!!

julsineya [31]

Answer:

39.6%

Step-by-step explanation: 99/250=39.6%

7 0

3 years ago

Read 2 more answers

Find the probability given that each set of events is independent: drawing a 3 from 5 cards numbered 1 through 5 and rolling an

ivanzaharov [21]

Answer:

1/10

Step-by-step explanation:

So first we need to find the probability of the cards, there is 5 cards and one card needs to be chosen, so the probability is 1/5

Then we need to find the probability of the dice, there is 3 even numbers in 1,2,3,4,5,and 6, so the probablity is 3/6 which is also equal to 1/2

Finaly we need to multiply 1/5 with 1/2, which is 1/10

6 0

4 years ago

Which is the equation of a line that has a slope of Negative two-thirds and passes through point (–3, –1)?