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2 years ago

Evaluate 2(s+t)^3-6when s = 3 and t = 2

Mathematics

1 answer:

Komok [63]2 years ago

3 0

Answer:

994

Step-by-step explanation:

2(s+t)= 10

10^3= 1.000

1.000-6= 994

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Dahasolnce [82]

Answer:

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Step-by-step explanation:

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2 years ago

PLEASE HELP TIMED

rusak2 [61]

A - drawing a red 8

B - drawing an 8

$P(A|B)=\dfrac{P(A\cap B)}{P(B)}\\\\ |\Omega|=52\\ |A\cap B|=2\\ |B|=4\\\\ P(A \cap B)=\dfrac{2}{52}=\dfrac{1}{26}\\ P(B)=\dfrac{4}{52}=\dfrac{1}{13}\\\\ P(A|B)=\dfrac{\dfrac{1}{26}}{\dfrac{1}{13}}=\dfrac{1}{26}\cdot13=\dfrac{1}{2}$

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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:

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

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

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

2 years ago

Please help. I will give brainliest!

Andrej [43]

Answer:

Step-by-step explanation:

How to Find Outliers Using the Interquartile Range(IQR)

Step 1: Find the IQR, Q1(25th percentile) and Q3(75th percentile). ...

Step 2: Multiply the IQR you found in Step 1 by 1.5: ...

Step 3: Add the amount you found in Step 2 to Q3 from Step 1: ...

Step 3: Subtract the amount you found in Step 2 from Q1 from Step 1:

7 0

2 years ago

Can somebody help me please

aleksklad [387]

Answer:

I think it's -2

Step-by-step explanation:

7 0

2 years ago

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