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katovenus [111]
3 years ago
14

A circular pond has a diameter of 4.5 feet.Which is closest to the circumference of the pond

Mathematics
2 answers:
ki77a [65]3 years ago
8 0
The circumference is the diameter times pi, so its 4.5 *<span> 3.141592653 
</span>=14.1371669
Snowcat [4.5K]3 years ago
3 0
Circumference = πD

Circumference of the pond = π(4.5) = 14.14 ft (nearest hundredth)

Answer: 14.14 ft
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X=14
3 times 14 is 42-13 is 29
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What is the ratio of red sections to total sections, in lowest terms?
Tanzania [10]

Hello there! Your answer would be 1/4.

So to start, count up the total number of sections, and the number of red sections. There are 16 sections total, and 4 of them are red.

This can give us the ratio 4/16.

But next we need to simplify. What is the biggest number that can be divided by both 4 and 16? 4. So, divide both 4 and 16 by 4 to get your simplified answer.

4÷4/16÷4

1/4

So, the ratio of red sections to total sections, in lowest terms is 1/4.

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3 years ago
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Find the volume of the solid.
dmitriy555 [2]

In Cartesian coordinates, the region (call it R) is the set

R = \left\{(x,y,z) ~:~ x\ge0 \text{ and } y\ge0 \text{ and } 2 \le z \le 4-x^2-y^2\right\}

In the plane z=2, we have

2 = 4 - x^2 - y^2 \implies x^2 + y^2 = 2 = \left(\sqrt2\right)^2

which is a circle with radius \sqrt2. Then we can better describe the solid by

R = \left\{(x,y,z) ~:~ 0 \le x \le \sqrt2 \text{ and } 0 \le y \le \sqrt{2 - x^2} \text{ and } 2 \le z \le 4 - x^2 - y^2 \right\}

so that the volume is

\displaystyle \iiint_R dV = \int_0^{\sqrt2} \int_0^{\sqrt{2-x^2}} \int_2^{4-x^2-y^2} dz \, dy \, dx

While doable, it's easier to compute the volume in cylindrical coordinates.

\begin{cases} x = r \cos(\theta) \\ y = r\sin(\theta) \\ z = \zeta \end{cases} \implies \begin{cases}x^2 + y^2 = r^2 \\ dV = r\,dr\,d\theta\,d\zeta\end{cases}

Then we can describe R in cylindrical coordinates by

R = \left\{(r,\theta,\zeta) ~:~ 0 \le r \le \sqrt2 \text{ and } 0 \le \theta \le\dfrac\pi2 \text{ and } 2 \le \zeta \le 4 - r^2\right\}

so that the volume is

\displaystyle \iiint_R dV = \int_0^{\pi/2} \int_0^{\sqrt2} \int_2^{4-r^2} r \, d\zeta \, dr \, d\theta \\\\ ~~~~~~~~ = \frac\pi2 \int_0^{\sqrt2} \int_2^{4-r^2} r \, d\zeta\,dr \\\\ ~~~~~~~~ = \frac\pi2 \int_0^{\sqrt2} r((4 - r^2) - 2) \, dr \\\\ ~~~~~~~~ = \frac\pi2 \int_0^{\sqrt2} (2r-r^3) \, dr \\\\ ~~~~~~~~ = \frac\pi2 \left(\left(\sqrt2\right)^2 - \frac{\left(\sqrt2\right)^4}4\right) = \boxed{\frac\pi2}

3 0
1 year ago
Can some one help me with this?
pychu [463]

Answer:

8 in

Step-by-step explanation:

Area = (diagonal 1 x diagonal 2)/2

60 = (diagonal 1 x 15)/2

60 x 2 = diagonal 1 x 15

120 = diagonal 1 x 15

diaognal 1 = 120/15 = 8

3 0
3 years ago
Find the taylor polynomial t3(x) for the function f centered at the number
inysia [295]

Answer:

t_3(x)=\frac{7\pi}{4}+\frac{7}{2}(x-1)-\frac{7}{4}(x-1)^2+\frac{7}{12}(x-1)^3

Step-by-step explanation:

We are given that

f(x)=7tan^{-1}(x)

a=1

T_n(x)=\sum_{r=0}^{n}\frac{f^r(a)(x-a)^r}{r!}

Substitute n=3 and a=1

t_3(x)=f(1)+f'(1)(x-1)+\frac{f''(1)(x-1)^2}{2!}+\frac{f'''(1)(x-1)^3}{3!}

f(x)=7tan^{-1}(x)

f(1)=7tan^{-1}(1)=7\times \frac{\pi}{4}=\frac{7\pi}{4}

Where tan^{-1}(1)=\frac{\pi}{4}

f'(x)=\frac{7}{1+x^2}

Using the formula

\frac{d(tan^{-1}(x))}{dx}=\frac{1}{1+x^2}

f'(1)=\frac{7}{2}

f''(x)=\frac{-14x}{(1+x^2)^2}

f''(1)=-\frac{7}{2}

f''(x)=-14x(x^2+1)^{-2}

f'''(x)=-14((x^2+1)^{-2}-4x^2(x^2+1)^{-3}})

By using the formula

(uv)'=u'v+v'u

f'''(x)=-14(\frac{x^2+1-4x^2}{(1+x^2)^3}

f'''(x)=(-14)\frac{-3x^2+1}{(1+x^2)^3}

f'''(1)=-14(\frac{-3(1)+1}{2^3})=\frac{7}{2}

Substitute the values

t_3(x)=\frac{7\pi}{4}+\frac{7}{2}(x-1)-\frac{7}{4}(x-1)^2+\frac{7}{2\times 3\times 2\times 1}(x-1)^3

t_3(x)=\frac{7\pi}{4}+\frac{7}{2}(x-1)-\frac{7}{4}(x-1)^2+\frac{7}{12}(x-1)^3

7 0
3 years ago
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