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Yuri [45]
3 years ago
8

Find the average value of f over region

Mathematics
1 answer:
yan [13]3 years ago
3 0
The area of D is given by:

\int\limits \int\limits {1} \, dA = \int\limits_0^7 \int\limits_0^{x^2} {1} \, dydx  \\  \\ = \int\limits^7_0 {x^2} \, dx =\left. \frac{x^3}{3} \right|_0^7= \frac{343}{3}

The average value of f over D is given by:

\frac{1}{ \frac{343}{3} }  \int\limits^7_0  \int\limits^{x^2}_0 {4x\sin(y)} \, dydx  = -\frac{3}{343}  \int\limits^7_0 {4x\cos(x^2)} \, dx  \\  \\ =-\frac{3}{343} \int\limits^{49}_0 {2\cos(t)} \, dt=-\frac{6}{343} \left[\sin(t)\right]_0^{49} \, dt=-\frac{6}{343}\sin49
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HELP ME PLEASE, I'D LIKE TO UNDERSTAND HOW TO DO THIS TOO SO I'LL GIVE BRAINLY FOR THE RIGHT ANSWER WITH GOOD AND UNDERSTANDING
umka2103 [35]

Answer:

34 goes in the blank.

Step-by-step explanation:

Triangle ABC is congruent to triangle TUV implies the following:

Angle A is congruent to angle T.

Angle B is congruent to angle U.

Angle C is congruent to angle V.

Side AB is congruent to side TU.

Side BC is congruent to side UV.

Side CA is congruent to side VT.

Note: Order tells you everything in a congruence statement.

Since side BC is congruent to UV, then the measure of side BC is equal to the measure of side UV.

8x+2=7x+6

Subtract 7x on both sides:

1x+2=6

x+2=6

Subtract 2 on both sides:

x=6-2

x=4

So if x=4, then the measure of side BC is 8x+2=8(4)+2=32+2=34.

So the measure of BC is 34 cm.

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4 years ago
Drew has two cats. One cat weighs 17 pounds, and the other one weighs 12 1/2 pounds. Audrey’s dog weighs 33 pounds. What is the
Step2247 [10]

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Step-by-step explanation:

17+12.5=29.5

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4 0
3 years ago
The projected rate of increase in enrollment at a new branch of the UT-system is estimated by E ′ (t) = 12000(t + 9)−3/2 where E
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Answer:

The projected enrollment is \lim_{t \to \infty} E(t)=10,000

Step-by-step explanation:

Consider the provided projected rate.

E'(t) = 12000(t + 9)^{\frac{-3}{2}}

Integrate the above function.

E(t) =\int 12000(t + 9)^{\frac{-3}{2}}dt

E(t) =-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+c

The initial enrollment is 2000, that means at t=0 the value of E(t)=2000.

2000=-\frac{24000}{\left(0+9\right)^{\frac{1}{2}}}+c

2000=-\frac{24000}{3}+c

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c=10,000

Therefore, E(t) =-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+10,000

Now we need to find \lim_{t \to \infty} E(t)

\lim_{t \to \infty} E(t)=-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+10,000

\lim_{t \to \infty} E(t)=10,000

Hence, the projected enrollment is \lim_{t \to \infty} E(t)=10,000

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3 years ago
The sum of two positive integers, a and b, is at least 30. The difference of the two integers is at least 10. If b is the greate
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