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barxatty [35]
3 years ago
10

HELP!! will give brainliest!

Mathematics
1 answer:
klemol [59]3 years ago
8 0

The rate of change of function A is greater than the rate of change of function B.

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6 because that is where the paarabola crosses the y line

Step-by-step explanation:

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Find the area A of the polygon with the given vertices. A(−6, 1), B(4, 1), C(4,−8), D(−6,−8)
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u must use the formula

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HELP HELP HELP HELP HELP
MA_775_DIABLO [31]

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I guess option A.is correct

4 0
3 years ago
Evaluate the surface integral ∫sf⋅ ds where f=⟨2x,−3z,3y⟩ and s is the part of the sphere x2 y2 z2=16 in the first octant, with
skad [1K]

Parameterize S by the vector function

\vec s(u,v) = \left\langle 4 \cos(u) \sin(v), 4 \sin(u) \sin(v), 4 \cos(v) \right\rangle

with 0 ≤ u ≤ π/2 and 0 ≤ v ≤ π/2.

Compute the outward-pointing normal vector to S :

\vec n = \dfrac{\partial\vec s}{\partial v} \times \dfrac{\partial \vec s}{\partial u} = \left\langle 16 \cos(u) \sin^2(v), 16 \sin(u) \sin^2(v), 16 \cos(v) \sin(v) \right\rangle

The integral of the field over S is then

\displaystyle \iint_S \vec f \cdot d\vec s = \int_0^{\frac\pi2} \int_0^{\frac\pi2} \vec f(\vec s) \cdot \vec n \, du \, dv

\displaystyle = \int_0^{\frac\pi2} \int_0^{\frac\pi2} \left\langle 8 \cos(u) \sin(v), -12 \cos(v), 12 \sin(u) \sin(v) \right\rangle \cdot \vec n \, du \, dv

\displaystyle = 128 \int_0^{\frac\pi2} \int_0^{\frac\pi2} \cos^2(u) \sin^3(v) \, du \, dv = \boxed{\frac{64\pi}3}

8 0
2 years ago
How can you write 0.07 in the form a and b
sp2606 [1]

Answer:

7/100

Step-by-step explanation:

We want to write .07 as a fraction

7/100  were a = 7 and b = 100

3 0
3 years ago
Read 2 more answers
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