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3 years ago

10

PLEASE HELP I WILL GIVE YOU POINTS,BRAINLIEST, AND A COOKIE !!

2 answers:

Anna007 [38]3 years ago

6 0

Answer:

20 maybe

Step-by-step explanation:

Marysya12 [62]3 years ago

6 0

Answer:

20 because you have to cross multiple

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A kitchen sink faucet streams 0.5 gallon of water in 10 seconds. Which faucet will fill a 3 gallon container faster?

SOVA2 [1]

60 secounds, or 1 minute

7 0

3 years ago

Evaluate f(7) *<br> f (x) = –2x² + 4

Firdavs [7]

Input 7 for every x so
f(7) = -2 (7)^2 +4
f(7) = -2(49) +4
f(7) = -98 +4
f(7) = -94

6 0

3 years ago

Read 2 more answers

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

if the slope is -3/5 and the y intercept is 1 what is the slope intercept form of the equation of each line given the slope and

Dennis_Churaev [7]

Answer:

y = -3/5x + 1

Step-by-step explanation:

3 0

2 years ago

Given that cos 63°≈ 0.454, enter the sine of a complementary angle.<br> sin

jek_recluse [69]

Answer:

$\cos(63^\circ)$ is the same as $\sin(27^\circ)$ by co-function identities

Step-by-step explanation:

Remember that complementary angles add up to 90°. The angle that i s complementary to 63° is 27°.

Also recall the co-function identities:

sin (90° – x) = cos x
cos (90° – x) = sin x

This means that $\cos(90^\circ-27^\circ)=\sin(27^\circ)\approx0.454$ .

4 0

1 year ago