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Sedbober [7]
3 years ago
12

Can someone help me please

Mathematics
2 answers:
vitfil [10]3 years ago
6 0

Answer:i would say the answer is A

Step-by-step explanation:its my personal opinion i hope it helps

rodikova [14]3 years ago
3 0
The answer is A just saying
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Wait a min i searched it up and it '=' to decimals. What does this equal to in fractions?
joja [24]
Answer: 12/55
Because: I did the math
3 0
3 years ago
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What is the following product? Assume d>/= 0
Hunter-Best [27]

Answer:

\boxed{\boxed{\sqrt[3]{d}\cdot \sqrt[3]{d}\cdot \sqrt[3]{d}=d}}

Step-by-step explanation:

The given expression is,

=\sqrt[3]{d}\cdot \sqrt[3]{d}\cdot \sqrt[3]{d}

It can also be written as,

=d^{\frac{1}{3}}\cdot d^{\frac{1}{3}}\cdot d^{\frac{1}{3}}

The exponent product rule of algebra states that, while multiplying two powers that have the same base, the exponents can be added.

As here all the terms have same base i.e d, so applying the rule

=d^{\frac{1}{3}+\frac{1}{3}+\frac{1}{3}}

=d^{\frac{1+1+1}{3}}

=d^{\frac{3}{3}}

=d^1

=d

5 0
3 years ago
Read 2 more answers
Help and show all work please
Anarel [89]

Answer:

1-3

2-8.9

3-6.3

4-26

5-7.8

6-17

7-10

8-32

9-7.9

10-13.2

11- is a right triangle

12- is not a right triangle

13- is a right triangle

14- is not a right triangle

15- is a right triangle

16- is not a right triangle

Step-by-step explanation:

I will do the rest it will just take me some time. I will also give you

step by step at the end.

7 0
3 years ago
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What is greater 0.13 or 13.5%?
Mumz [18]

Answer:13.5% is greater than 0.13

Step-by-step explanation:

0.13 or 13.5% which is greater

13.5%=13.5/100=0.135

0.13 or 0.135 which is greater

0.135 is greater which means 13.5% is greater

3 0
3 years ago
Find the outward flux of the vector field f=(x3,y3,z2) across the surface of the region that is enclosed by the circular cylinde
ahrayia [7]
Use the divergence theorem.

\mathbf f(x,y,z)=(x^3,y^3,z^3)\implies \nabla\cdot\mathbf f=3(x^2+y^2+z^2)

Calling the closed region \mathcal R and its boundary the surface \partial\mathcal R, we have by the divergence theorem,

\displaystyle\underbrace{\iint_{\partial\mathcal R}\mathbf f\cdot\mathrm d\mathbf S}_{\text{flux}}=\iiint_{\mathcal R}\nabla\cdot\mathbf f\,\mathrm dV

We set up the volume integral in cylindrical coordinates, using

\begin{cases}x=u\cos v\\y=u\sin v\\z=w\end{cases}\implies\mathrm dV=u\,\mathrm du\,\mathrm dv\,\mathrm dw

\displaystyle\iiint_{\mathcal R}\nabla\cdot\mathbf f\,\mathrm dV=3\int_{w=0}^{w=6}\int_{v=0}^{v=2\pi}\int_{u=0}^{u=8}(u^2\cos^2v+u^2\sin^2v+w^2)u\,\mathrm du\,\mathrm dv\,\mathrm dw
=6\pi\displaystyle\int_{w=0}^{w=6}\int_{u=0}^{u=8}(u^3+uw^2)\,\mathrm du\,\mathrm dw=50,688\pi
7 0
3 years ago
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