All categories

3 years ago

Need help !!!!!!!!!!!!!!!!!!!

Mathematics

1 answer:

Lesechka [4]3 years ago

8 0

Answer:

I'm guessing the answer is B againn XD

must be a lucky day for B!!

You might be interested in

Joe Morris Plains to fertilize his 1800 foot wide Hayfield he will be using his truck spreader that spreads of 60-foot strip how

gayaneshka [121]

Answer:

He should do 30 strips

Step-by-step explanation:60 times 3 is180 so and aa 0 to 180 and get 1800 so then you and the 0 to 3 and get 30

6 0

3 years ago

The area of a circle is 81 ft 2 . What is the π circumference? Leave in terms of Pi.

KonstantinChe [14]

Answer:

10.2π

Step-by-step explanation:

Given data

Area of circle= 81 ft^2

Also, the expression for the area of a circle is

A=πr^2

substitute

81=3.142*r^2

r^2= 81/3.142

r^2= 25.77

square both sides

r= √25.77

r= 5.1

Also, the expression for the circumference is

C= 2πr

substitute

C= 2*π*5.1

C= 10.2π

Hence the circumference is 10.2π

7 0

2 years ago

Use the given transformation x=4u, y=3v to evaluate the integral. ∬r4x2 da, where r is the region bounded by the ellipse x216 y2

exis [7]

The Jacobian for this transformation is

$J = \begin{bmatrix} x_u & x_v \\ y_u & y_v \end{bmatrix} = \begin{bmatrix} 4 & 0 \\ 0 & 3 \end{bmatrix}$

with determinant $|J| = 12$ , hence the area element becomes

$dA = dx\,dy = 12 \, du\,dv$

Then the integral becomes

$\displaystyle \iint_{R'} 4x^2 \, dA = 768 \iint_R u^2 \, du \, dv$

where $R'$ is the unit circle,

$\dfrac{x^2}{16} + \dfrac{y^2}9 = \dfrac{(4u^2)}{16} + \dfrac{(3v)^2}9 = u^2 + v^2 = 1$

so that

$\displaystyle 768 \iint_R u^2 \, du \, dv = 768 \int_{-1}^1 \int_{-\sqrt{1-v^2}}^{\sqrt{1-v^2}} u^2 \, du \, dv$

Now you could evaluate the integral as-is, but it's really much easier to do if we convert to polar coordinates.

$\begin{cases} u = r\cos(\theta) \\ v = r\sin(\theta) \\ u^2+v^2 = r^2\\ du\,dv = r\,dr\,d\theta\end{cases}$

Then

$\displaystyle 768 \int_{-1}^1 \int_{-\sqrt{1-v^2}}^{\sqrt{1-v^2}} u^2\,du\,dv = 768 \int_0^{2\pi} \int_0^1 (r\cos(\theta))^2 r\,dr\,d\theta \\\\ ~~~~~~~~~~~~ = 768 \left(\int_0^{2\pi} \cos^2(\theta)\,d\theta\right) \left(\int_0^1 r^3\,dr\right) = \boxed{192\pi}$