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AfilCa [17]
3 years ago
11

Which expression has a value of 20? O (16 – 8) x 2 +4 (16 – 8) x (2+4) 16 - 8 X (2+4) O 16 — (8 x 2 + 4)​

Mathematics
1 answer:
Paha777 [63]3 years ago
4 0

Answer:

(16-8)*(2+4) is the correct answer

SO it's A

Step-by-step explanation:

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If ∫−1−4f(x)dx=0 and ∫31g(x)dx=3, what is the value of ∫∫Df(x)g(y)dA where D is the square: −4≤x≤−1, 1≤y≤3?
Lana71 [14]

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0

Step-by-step explanation:

From your problem, we have to extract the information that are important from the first two intregrals so we can solve the double integral.

\int\limits^{-1}_{-4} {f(x)} \, dx = 0

We also have that:

\int\limits^{3}_{1} {g(x)} \, dx = 3

---------------------------------

With this, now we can solve the double integral.

Since the limits of integration are constant, i can use dA both as dydx or dxdy. I am going to use dydx.

So the double integral will be:

\int \limits^{-1}_{-4} \int \limits^{3}_{1} {g(y)} {f(x)} dy dx\

We solve a double integral from the inside to the outside, so the first integral we solve is:

\int \limits^{3}_{1} \y g(y) \x f(x) dy

f is a function of x and we are integrating dy, so this means that f is a constant. Our integral now is this:

\x f(x) \int \limits^{3}_{1} \y g(y) dy

From above, we have that

\int \limits^{3}_{1} \x g(x) dx = 3

So,

\int \limits^{3}_{1} \y g(y) dy = 3

Now we have to solve the outside integral:

3\int\limits^{-1}_{-4} {f(x)} \, dx

We know that

\int\limits^{-1}_{-4} {f(x)} \, dx = 0

So the double integral will be 0

3 0
3 years ago
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