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Debora [2.8K]
2 years ago
5

Is 94,083,392 divisible by 8? HELP ME PLZ!!!

Mathematics
2 answers:
morpeh [17]2 years ago
7 0

Answer:

no it divisible by 3

Step-by-step explanation:

Marina86 [1]2 years ago
7 0

Answer:

No. Hope this helps you :)

Step-by-step explanation:

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For the monthly rainfall data set of 7 cm, 3 cm, 10 cm, 7 cm, 12 cm, 10 cm, 8 cm, and 7 cm calculate the mean. * what is the mea
timama [110]
The mean would be 8 since you would have to add all of them up which would give you 64 and then divide by the amount of numbers their are which is 8 so 64/8 = 8
5 0
3 years ago
Click an item in the list or group of pictures at the bottom of the problem and, holding the button down, drag it into the corre
Scrat [10]

Answer:

4√(xy³)

Step-by-step explanation:

8√(x²y⁶)

The above expression can be simplified as follow:

8√(x²y⁶)

Recall:

m√a = a^1/m

Therefore,

8√(x²y⁶) = (x²y⁶)^1/8

Recall:

(aⁿ)^1/m = a^(n/m)

Therefore,

(x²y⁶)^1/8 = x^(2/8)•y^(6/8)

= x^1/4•y^3/4

= (xy³)^1/4

Recall :

a^1/m = m√a

Therefore,

(xy³)^1/4 = 4√(xy³)

Therefore,

8√(x²y⁶) = 4√(xy³)

6 0
3 years ago
Read 2 more answers
For questions 3 that’s all i need help please !!! For a test and please put your answers below !!!
garik1379 [7]

Answer:

D) 1078

Step-by-step explanation:

4 0
3 years 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
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