Ask question

Ask question

All categories

3 years ago

7

Evaluate the following double integral: xy dA D where the region D is the triangular region whose vertices are (0, 0), (0, 3), (

3, 0).

1 answer:

natulia [17]3 years ago

4 0

Answer:

I= 84

Step-by-step explanation:

for

$I=\int\limits^{}_{} \int\limits^{}_D {x*y} \, dA = \int\limits^{}_{} \int\limits^{}_D {x*y} \, dx*dy$

since D is the rectangle such that 0<x<3 , 0<y<3

$I=\int\limits^{}_{} \int\limits^{}_D {x*y} \, dA = \int\limits^{3}_{0} \int\limits^{3}_{0} {x*y} \, dx*dy = \int\limits^{3}_{0} {x} \, dx\int\limits^{3}_{0} {y} \, dy = x^{2} /2*y^{2} /2 = (3^{2} /2 - 0^{2} /2)* (3^{2} /2 - 0^{2} /2) = 3^{4} /4 = 81/4$

You might be interested in

Jonah drew two squares with the same dimension. He then added 2 inches to the length of one square to make it a rectangle. He al

Lubov Fominskaja [6]

Answer:

Perimeter of rectangle added 2 to length:

perimeter of rectangle= length + width + length + width

P = ( l + 2 ) + w + ( l + 2 ) + w

P = 2 ( l + 2 ) + 2 w

P = 4 + 2 l + 2 w

Perimeter of rectangle added 2 to width:

P = l + w + l + w

p = l + ( w + 2) + l + ( w + 2)

p = 2 l + 2 ( w + 2)

p = 2 l + 2 w + 4

5 0

4 years ago

CAN ANYBODY GOOD WITH MATH HELP ME OUT PLEASE.

inna [77]

Answer:

Just put the money divided by the percent in a calculator-

Step-by-step explanation:

:/

4 0

3 years ago

Can someone please explain how to do this.

Cerrena [4.2K]

Well you can graph it by plotting 3 points then drawing a line through these 3 points. The graph will be a straight line . By plotting 3 points you can be more sure if you are right, because if you make a mistake then they might not make a straight line.

Put y = 0 in the equation and find the x coordinate:-

x + 5(0) = -20

x = -20

So we have one point to plot:- (-20,0)

Putting x = 0 we get 5y = -20 so y = -4

Second point:- (0, -4)

Putting x = 5 say we get 4 + 5y = -20 so y = -25/5 = -5

so our 3rd point is (5, -5)

7 0

4 years ago

Read 2 more answers

The function f(x) = 60x + 4 represents the amount of money Mike earns making custom tables. The function g(x) = 2x - 4 represent

siniylev [52]

Answer:

$244

Step-by-step explanation:

f(x) = 60x + 4 and g(x) = 2x-4

We are told to find f(g=4), when Mike makes four tables.

We have to put replace x in f(x) with the value g(x) when x is 4, like this:

f(x) = 60x + 4 // Replace x with g(x) = 2x-4

f(x) = 60(2x-4) + 4

f(x) = (120x - 240) + 4 // Solve

f(x) = (120*4 - 240) +4 // They tell us x=4

f(x) = 480 - 240 + 4 = $244

3 0

3 years ago

Read 2 more answers

Question Help Suppose that the lifetimes of light bulbs are approximately normally distributed, with a mean of 5656 hours and a

koban [17]

Answer:

a)3.438% of the light bulbs will last more than 6262 hours.

b)11.31% of the light bulbs will last 5252 hours or less.

c) 23.655% of the light bulbs are going to last between 5858 and 6262 hours.

d) 0.12% of the light bulbs will last 4646 hours or less.

Step-by-step explanation:

Normally distributed problems can be solved by the z-score formula:

On a normaly distributed set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a value X is given by:

$Z = \frac{X - \mu}{\sigma}$

After we find the value of Z, we look into the z-score table and find the equivalent p-value of this score. This is the probability that a score will be LOWER than the value of X.

In this problem, we have that:

The lifetimes of light bulbs are approximately normally distributed, with a mean of 5656 hours and a standard deviation of 333.3 hours.

So $\mu = 5656, \sigma = 333.3$

(a) What proportion of light bulbs will last more than 6262 hours?

The pvalue of the z-score of $X = 6262$ is the proportion of light bulbs that will last less than 6262. Subtracting 100% by this value, we find the proportion of light bulbs that will last more than 6262 hours.

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{6262 - 5656}{333.3}$

$Z = 1.82$

$Z = 1.81$ has a pvalue of .96562. This means that 96.562% of the light bulbs are going to last less than 6262 hours. So

$P = 100% - 96.562% = 3.438%$ of the light bulbs will last more than 6262 hours.

(b) What proportion of light bulbs will last 5252 hours or less?

This is the pvalue of the zscore of $X = 5252$

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{5252- 5656}{333.3}$

$Z = -1.21$

$Z = -1.21$ has a pvalue of .1131. This means that 11.31% of the light bulbs will last 5252 hours or less.

(c) What proportion of light bulbs will last between 5858 and 6262 hours?

This is the pvalue of the zscore of $X = 6262$ subtracted by the pvalue of the zscore $X = 5858$

For $X = 6262$ , we have that $Z = 1.81$ with a pvalue of .96562.

For X = 5858

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{5858- 5656}{333.3}$

$Z = 0.61$

$Z = 0.61$ has a pvalue of .72907.

So, the proportion of light bulbs that will last between 5858 and 6262 hours is

$P = .96562 - .72907 = .23655$

23.655% of the light bulbs are going to last between 5858 and 6262 hours.

(d) What is the probability that a randomly selected light bulb lasts less than 4646 hours?

This is the pvalue of the zscore of $X = 4646$

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{4646- 5656}{333.3}$

$Z = -3.03$

$Z = -3.03$ has a pvalue of .0012. This means that 0.12% of the light bulbs will last 4646 hours or less.

5 0

3 years ago