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2 years ago

Elan is painting the outside of a rectangular barn door. It is

Mathematics

1 answer:

UkoKoshka [18]2 years ago

5 0

Answer:

9600

Step-by-step explanation:

80 times 60 times 2

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The ages of Jamal, Wesley, an John are consecutive integers.the sum of thier ages is 108.what are thier ages .

hodyreva [135]

Answer:

Step-by-step explanation:

We have three people

jamal = x (since im guessing hes the youngest)

wesley= x+1

John= x+2

the formula you would use is x+x+1+x+2=108

when you simplify that you get 3X+3=108

you subtract the 3 from 108 which leaves you with 3X=105

divide that by 3 which makes x=35

jamal is 35

wesley is 36

john is 37

3 0

2 years ago

PLS HELP ME ASAP FOR 12 (SHOW WORK!!)

alex41 [277]

160 is the first because 1 inches 20 but it says 2 inches so at 40+40+40+40
and the 2nd part is yes he is correct because he converted the inches into feet so i think he is also correct

6 0

2 years ago

Solve your equations below and right the algebraic steps, n/12 = 10

sdas [7]

$\frac{n}{12}=10$

Multiply both sides of the equation by 12

$\frac{n}{12}\times12=10\times12$ $n=\text{ 120}$

7 0

11 months 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

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

3 0

1 year ago

Find the perimeter of the triangle .

lawyer [7]

Answer:

P = 9x + 22

Step-by-step explanation:

Perimeter = sum of all sides

P = (3x + 11) + (3x + 11) + (3x)

P = 9x + 22

I hope this helps! God bless

5 0

1 year ago