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Julli [10]
3 years ago
9

F( y ) = y + y 2 - 3

Mathematics
1 answer:
ExtremeBDS [4]3 years ago
7 0
F(y) = y + y^2 - 3

f(-2) = -2 + (-2^2) - 3
f(-2) = -2 + 4 - 3
f(-2) = -1

f(-4) = -4 + (-4^2) - 3
f(-4) = -4 + 16 - 3
f(-4) = 9

f(0) = 0 + 0^2 - 3
f(0) = -3

f(2) = 2 + 2^2 - 3
f(2) = 2 + 4 - 3
f(2) = 3

f(4) = 4 + 4^2 - 3
f(4) = 4 + 16 - 3
f(4) = 17



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2x + 3y = 3 and x + 6y = -3
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Answer:

Step-by-step explanation:

Multiplying x+6y=-3 with 2 so its equal to first eqn,

Eliminating,

2x + 3y = 3        (1)

2x +12y = -6       (2)                  (because subtracting)

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

      -9y =9

     ∴y=-1

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Find the average value of f over region
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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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cupoosta [38]

Answer:

C is the answer.

Step-by-step explanation:

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sveticcg [70]

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Step-by-step explanation:

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Step-by-step explanation:

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9+27= 36

7 0
3 years ago
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