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mario62 [17]
2 years ago
5

The radius of a circle is 9 inches. What is the circle's area? Use 3.14 for ​.

Mathematics
2 answers:
Anna007 [38]2 years ago
5 0

Answer:

<u>254.34 in²</u>

Step-by-step explanation:

<u>Area of a circle</u>

  • A = πr²
  1. <u>r = radius of the circle</u>

<u>Solving</u>

  • A = 3.14 x (9)²
  • A = 3.14 x 81
  • A = <u>254.34 in²</u>
slamgirl [31]2 years ago
5 0

Answer:

254.47 in²(rounded to nearest hundredth)

_________________________________________________________

A = \pi r^2

<em>tiny r means radius</em>

<em>² means multiply the number by itself twice</em>

<em />\pi r^2 = \pi * 9^2 = 81\pi(254.47~in^2)

_________________________________________________________

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A circle has a diameter of 4.5 inches what is the circumference​
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Answer:

c = 4.5π inches   exact answer

c = 14.13 inches   decimal approximation

Step-by-step explanation:

c = πd

c = 4.5π    exact answer

using 3.14 for π

c = 3.14 * 4.5

c = 14.13    decimal approximation

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2 years ago
A negative number divided by a positive number will yield a<br>number.​
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Answer: negative Number

Step-by-step explanation:

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3 years ago
In this problem, you will use undetermined coefficients to solve the nonhomogeneous equation y′′+4y′+4y=12te^(−2t)−(8t+12) with
Zarrin [17]

First check the characteristic solution:

<em>y''</em> + 4<em>y'</em> + 4<em>y</em> = 0

has characteristic equation

<em>r</em> ² + 4<em>r</em> + 4 = (<em>r</em> + 2)² = 0

with a double root at <em>r</em> = -2, so the characteristic solution is

y_c = C_1e^{-2t} + C_2te^{-2t}

For the particular solution corresponding to 12te^{-2t}, we might first try the <em>ansatz</em>

y_p = (At+B)e^{-2t}

but e^{-2t} and te^{-2t} are already accounted for in the characteristic solution. So we instead use

y_p = (At^3+Bt^2)e^{-2t}

which has derivatives

{y_p}' = (-2At^3+(3A-2B)t^2+2Bt)e^{-2t}

{y_p}'' = (4At^3+(-12A+4B)t^2+(6A-8B)t+2B)e^{-2t}

Substituting these into the left side of the ODE gives

(4At^3+(-12A+4B)t^2+(6A-8B)t+2B)e^{-2t} + 4(-2At^3+(3A-2B)t^2+2Bt)e^{-2t} + 4(At^3+Bt^2)e^{-2t} \\\\ = (6At+2B)e^{-2t} = 12te^{-2t}

so that 6<em>A</em> = 12 and 2<em>B</em> = 0, or <em>A</em> = 2 and <em>B</em> = 0.

For the second solution corresponding to -8t-12, we use

y_p = Ct + D

with derivative

{y_p}' = C

{y_p}'' = 0

Substituting these gives

4C + 4(Ct+D) = 4Ct + 4C + 4D = -8t-12

so that 4<em>C</em> = -8 and 4<em>C</em> + 4<em>D</em> = -12, or <em>C</em> = -2 and <em>D</em> = -1.

Then the general solution to the ODE is

y = C_1e^{-2t} + C_2te^{-2t} + 2t^3e^{-2t} - 2t - 1

Given the initial conditions <em>y</em> (0) = -2 and <em>y'</em> (0) = 1, we have

-2 = C_1 - 1 \implies C_1 = -1

1 = -2C_1 + C_2 - 2 \implies C_2 = 1

and so the particular solution satisfying these conditions is

y = -e^{-2t} + te^{-2t} + 2t^3e^{-2t} - 2t - 1

or

\boxed{y = (2t^3+t-1)e^{-2t} - 2t - 1}

7 0
2 years ago
I believe this is goodbye i hope i helped u guys.
jeyben [28]

Answer:

goodbye??? what're you going to bed? you alr my dude?

Step-by-step explanation:

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2 years ago
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Juliette [100K]

Answer:

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7 0
3 years ago
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