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liberstina [14]
2 years ago
11

Hi I don't know if im putting the answer in wrong but it keeps saying its wrong for the rate of change question but for the slop

e question it keeps saying its wrong and the slope has to be in fraction form but its not 6/4 or 9/8 it has to be a negative number but I can't seem to figure these out please help this is 244 points of my grade

Mathematics
1 answer:
Varvara68 [4.7K]2 years ago
7 0
Ok ur answer is 35 :) hope this helps (:)(:)(:)(:)(:)(:)(:)(:)(:)
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patriot [66]

Answer:

8 is the correct answer because coefficient lies before Exponent. Exponent is variable

So 8 is the coefficient.

5 0
2 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
Suppose that receiving stations​ X, Y, and Z are located on a coordinate plane at the points ​(6​,2​), ​(negative 2​,negative 4​
marin [14]

Answer:

  (2,-1)

Step-by-step explanation:

A graph is useful here. Points X and Y have coordinate differences of ...

  X -Y = (6, 2) -(-2, -4) = (6+2, 2+4) = (8, 6)

Then the distance between X and Y is ...

  d = √(8² +6²) = √100 = 10

The point 5 units from X and from Y is the midpoint of XY:

  E = (X +Y)/2 = ((6, 2) +(-2, -4))/2 = (4, -2)/2 = (2, -1)

The epicenter is (2, -1).

_____

The graph shows circles of radius 5 around X and Y, and a circle of radius 13 around Z. The circles intersect at the point (2, -1), the epicenter.

5 0
3 years ago
Write the equation of the line that passes through the points (3, 6) and (5, 18) using function notation.
faust18 [17]

The answer would be B: F(x) = 6x - 12

This line passes through both given points.

6 0
3 years ago
Read 2 more answers
What is the solution to -|-5 + 2|.
prisoha [69]
Im guessing those are those brackets?
3 0
3 years ago
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