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valentinak56 [21]
3 years ago
11

A cylinder hot tub with a diameter of 12ft and a height of 6ft is partially with 500ft^3 of water what is the volume? AND how ma

ny cubic ft of water can it hold
Mathematics
1 answer:
Ugo [173]3 years ago
4 0
Idk idk idk idk idk idk idk idk idk idk idk
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A basketball player scored 26 points in one game. In basketball, some baskets are worth 3 points, some are worth 2 points, and f
Vsevolod [243]

Answer:

3 points = 2

2 points = 6

1 point = 8

Step-by-step explanation:

Given that:

Total point scored = 26

Let number of 3 point basket = x

Number of 2 point basket = x + 4

Number of free throws = x + x + 4 = 2x + 4

Hence,

Total 3 points = 3x

Total 2 points 2(x + 4)

Total free throw points = 2x + 4

3x + 2(x + 4) + 2x + 4 = 26

3x + 2x + 8 + 2x + 4 = 26

7x + 12 = 26

7x = 26 - 12

7x = 14

x = 2

Number of 3 points = x = 2

2 points = (x+4) = 2+4 = 6

1 point = (2x+ 4) = 2(2) + 4 = 8

5 0
3 years ago
Read 2 more answers
Help asap !………………………
My name is Ann [436]

Answer:

c

Step-by-step explanation:

dvdywuwhfuwi eurue wuid

4 0
1 year ago
Choose the best coordinate system to find the volume of the portion of the solid sphere rho <_4 that lies between the cones φ
MrRissso [65]

Answer:

So,  the volume is:

\boxed{V=\frac{128\sqrt{2}\pi}{3}}

Step-by-step explanation:

We get the limits of integration:

R=\left\lbrace(\rho, \varphi, \theta):\, 0\leq \rho \leq  4,\, \frac{\pi}{4}\leq \varphi\leq \frac{3\pi}{4},\, 0\leq \theta \leq 2\pi\right\rbrace

We use the spherical coordinates and  we calculate a triple integral:

V=\int_0^{2\pi}\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}}\int_0^4  \rho^2 \sin \varphi \, d\rho\, d\varphi\, d\theta\\\\V=\int_0^{2\pi}\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \sin \varphi \left[\frac{\rho^3}{3}\right]_0^4\, d\varphi\, d\theta\\\\V=\int_0^{2\pi}\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \sin \varphi \cdot \frac{64}{3} \, d\varphi\, d\theta\\\\V=\frac{64}{3} \int_0^{2\pi} [-\cos \varphi]_{\frac{\pi}{4}}^{\frac{3\pi}{4}}  \, d\theta\\\\V=\frac{64}{3} \int_0^{2\pi} \sqrt{2} \, d\theta\\\\

we get:

V=\frac{64}{3} \int_0^{2\pi} \sqrt{2} \, d\theta\\\\V=\frac{64\sqrt{2}}{3}\cdot[\theta]_0^{2\pi}\\\\V=\frac{128\sqrt{2}\pi}{3}

So,  the volume is:

\boxed{V=\frac{128\sqrt{2}\pi}{3}}

4 0
3 years ago
How can i rewrite that with exponents instead of repeated multiplication
Montano1993 [528]

Answer:

10^12

Step-by-step explanation:

Add the exponents when you multiply numbers with exponents

8 0
3 years ago
I need help please!!!
frutty [35]
The correct answer is B: f=4.5n
5 0
2 years ago
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