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2 years ago

6

there's a 100 dollar stuck up 12 feet up a building and you buy a 15 foot ladder that can be safely leaned at a 60 degree angle,

can you reach the bill?

1 answer:

maks197457 [2]2 years ago

7 0

No i cannot be safely leaned

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Name the fourth in pane wxv

snow_tiger [21]

Your answer would be U

8 0

3 years ago

The circle below is centered at (4, -1) and has a radius of 3. What is its

denis23 [38]

Answer:

Answers C and D are proably missing the y term. I'll assume they are:

C. (x-4)² + (y-1)² = 9

D. (x-4)² + (y-1)² = 3

A, is the equation that produces the drawn circle.

Step-by-step explanation:

See the attached image for explanation.

6 0

2 years ago

Find the mass of the lamina that occupies the region D = {(x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1} with the density function ρ(x, y) = xye

Alona [7]

Answer:

The mass of the lamina is 1

Step-by-step explanation:

Let $\rho(x,y)$ be a continuous density function of a lamina in the plane region D,then the mass of the lamina is given by:

$m=\int\limits \int\limits_D \rho(x,y) \, dA$ .

From the question, the given density function is $\rho (x,y)=xye^{x+y}$ .

Again, the lamina occupies a rectangular region: D={(x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1}.

The mass of the lamina can be found by evaluating the double integral:

$I=\int\limits^1_0\int\limits^1_0xye^{x+y}dydx$ .

Since D is a rectangular region, we can apply Fubini's Theorem to get:

$I=\int\limits^1_0(\int\limits^1_0xye^{x+y}dy)dx$ .

Let the inner integral be: $I_0=\int\limits^1_0xye^{x+y}dy$ , then

$I=\int\limits^1_0(I_0)dx$ .

The inner integral is evaluated using integration by parts.

Let $u=xy$ , the partial derivative of u wrt y is

$\implies du=xdy$

and

$dv=\int\limits e^{x+y} dy$ , integrating wrt y, we obtain

$v=\int\limits e^{x+y}$

Recall the integration by parts formula: $\int\limits udv=uv- \int\limits vdu$

This implies that:

$\int\limits xye^{x+y}dy=xye^{x+y}-\int\limits e^{x+y}\cdot xdy$

$\int\limits xye^{x+y}dy=xye^{x+y}-xe^{x+y}$

$I_0=\int\limits^1_0 xye^{x+y}dy$

We substitute the limits of integration and evaluate to get:

$I_0=xe^x$

This implies that:

$I=\int\limits^1_0(xe^x)dx$ .

Or

$I=\int\limits^1_0xe^xdx$ .

We again apply integration by parts formula to get:

$\int\limits xe^xdx=e^x(x-1)$ .

$I=\int\limits^1_0xe^xdx=e^1(1-1)-e^0(0-1)$ .

$I=\int\limits^1_0xe^xdx=0-1(0-1)$ .

$I=\int\limits^1_0xe^xdx=0-1(-1)=1$ .

No unit is given, therefore the mass of the lamina is 1.

3 0

3 years ago

Please help DO NOT IGNORE DO NOT IGNORE

lisov135 [29]

Answer:

24.997 cm

Step-by-step explanation:

Find the diameter:

7(2) = 14 cm

Find the circumference:

C = $\pi d$

C = $\pi (14)$

C = 43.988

Divide the circumference by 4:

43.988/4 = 10.997

(This is the length of the curved side)

Add length to the other 2 sides:

7 + 7 + 10.997 = 24.997

3 0

3 years ago

DaniilM [7]

g) c=1.98w is your answer

w = weight = 5.5

c = cost = 10.89

10.89 = 1.98(5.5)

10.89 = 10.89 (true)

hope this helps

8 0

3 years ago