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2 years ago

What is the nth term of this quadratic sequence: 3,11,25,45

1 answer:

6 0

<span>This solution might be a bit out there, but it should probably work. For this, we will consider sequencesto have an invisible "0th" term to make calculation easier.</span>Missing: <span>quadratic</span>

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Find the area of the shaded region.

lbvjy [14]

Answer:

$48\pi\\\approx 150.796$

Step-by-step explanation:

1.Approach

To solve this problem, find the area of the larger circle, and the area of the smaller circle. Then subtract the area of the smaller circle from the larger circle to find the area of the shaded region.

2.Find the area of the larger circle

The formula to find the area of a circle is the following,

$A=(\pi)(r^2)$

Where (r) is the radius, the distance from the center of the circle to the circumference, the outer edge of the circle. ( $\pi$ ) represents the numerical constant (3.1415...). One is given that the radius of (8), substitute this into the formula and solve for the area,

$A_l=(\pi)(r^2)\\A_l=(\pi)(8^2)\\A_l=(\pi)(64)$

3.Find the area of the smaller circle

To find the area of the smaller circle, one must use a very similar technique. One is given the diameter, the distance from one end to the opposite end of a circle. Divide this by two to find the radius of the circle.

8 ÷2 = 4

Radius = 4

Substitute into the formula,

$A_s=(\pi)(r^2)\\A_s=(\pi)(4^2)\\A_s=(\pi)(16)$

4.Find the area of the shaded region

Subtract the area of the smaller circle from the area of the larger circle.

$A_l-A_s\\=64\pi - 16\pi\\=48\pi$

$\approx150.796$

5 0

2 years ago

How much has James earned since the beginning of the year in 2014? (use pay stub) *

schepotkina [342]

Can you give more information?

4 0

2 years ago

Evaluate the triple integral ∭EzdV where E is the solid bounded by the cylinder y2+z2=81 and the planes x=0,y=9x and z=0 in the

dem82 [27]

Answer:

I = 91.125

Step-by-step explanation:

Given that:

$I = \int \int_E \int zdV$ where E is bounded by the cylinder $y^2 + z^2 = 81$ and the planes x = 0 , y = 9x and z = 0 in the first octant.

The initial activity to carry out is to determine the limits of the region

since curve z = 0 and $y^2 + z^2 = 81$

∴ $z^2 = 81 - y^2$

$z = \sqrt{81 - y^2}$

Thus, z lies between 0 to $\sqrt{81 - y^2}$

GIven curve x = 0 and y = 9x

$x =\dfrac{y}{9}$

As such,x lies between 0 to $\dfrac{y}{9}$

Given curve x = 0 , $x =\dfrac{y}{9}$ and z = 0, $y^2 + z^2 = 81$

y = 0 and

$y^2 = 81 \\ \\ y = \sqrt{81} \\ \\ y = 9$

∴ y lies between 0 and 9

Then $I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \int^{\sqrt{81-y^2}}_{z=0} \ zdzdxdy$

$I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix} \dfrac{z^2}{2} \end {bmatrix} ^ {\sqrt {{81-y^2}}}_{0} \ dxdy$

$I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix} \dfrac{(\sqrt{81 -y^2})^2 }{2}-0 \end {bmatrix} \ dxdy$

$I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix} \dfrac{{81 -y^2} }{2} \end {bmatrix} \ dxdy$

$I = \int^9_{y=0} \begin {bmatrix} \dfrac{{81x -xy^2} }{2} \end {bmatrix} ^{\dfrac{y}{9}}_{0} \ dy$

$I = \int^9_{y=0} \begin {bmatrix} \dfrac{{81(\dfrac{y}{9}) -(\dfrac{y}{9})y^2} }{2}-0 \end {bmatrix} \ dy$

$I = \int^9_{y=0} \begin {bmatrix} \dfrac{{81 \ y -y^3} }{18} \end {bmatrix} \ dy$

$I = \dfrac{1}{18} \int^9_{y=0} \begin {bmatrix} {81 \ y -y^3} \end {bmatrix} \ dy$

$I = \dfrac{1}{18} \begin {bmatrix} {81 \ \dfrac{y^2}{2} - \dfrac{y^4}{4}} \end {bmatrix}^9_0$

$I = \dfrac{1}{18} \begin {bmatrix} {40.5 \ (9^2) - \dfrac{9^4}{4}} \end {bmatrix}$

$I = \dfrac{1}{18} \begin {bmatrix} 3280.5 - 1640.25 \end {bmatrix}$

$I = \dfrac{1}{18} \begin {bmatrix} 1640.25 \end {bmatrix}$

I = 91.125

4 0

2 years ago

It rained 10% of the 30 days last month.on how many days did it not rain?

drek231 [11]

10% of 30 is 3, which means it rained 3 days.

30-3 = 27 days NO rain

7 0

2 years ago

Read 2 more answers

Using the distributive property,simplify the following: 5(x+2)

MArishka [77]

The answer is 5x + 10 because you have to times 5 by everything in the parenteses

3 0

3 years ago