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3 years ago

If the diameter of his flower bed is 100 inches what is distance around the garden

1 answer:

8 0

Well if pi is referred to as 3.14 then the distance around the garden would be 314.
The formula for Circumfrence is pi times diameter, so 3.14 times 100 = 314 (circumfrence)

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Answer pls thanks kfdsdatttdfyui8u90g

diamong [38]

Answer:

Step-by-step explanation:

960 75%

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2 years ago

Jessica works between 12h and 18h per week at her part-time job. She earns $9.25 per hour. The function p(h) = 9.25h models the

Tpy6a [65]

D. All real numbers between 111 and 166.5 inclusive.

3 0

3 years ago

Kelsey has already spent 42 minutes on the phone, and she expects to spend 2 more minutes

grandymaker [24]

Answer:

11 calls

Step-by-step explanation:

She has already spent 42 min. Also she wants to talk for 2 min per call. We can change calls on x. So, only 2x min for her calls left.

42 + 2x = 64

2x = 64 - 42

2x= 22

x=11

Answer: 11 calls

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3 years ago

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

Need by TONIGHT please help

igor_vitrenko [27]

Slope = 1/2 :)
rise over run (rise/run)
so... 2/4 = 1/2
hope this helps!

7 0

3 years ago