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3 years ago

Help!!!! I don't get this

Mathematics

2 answers:

Serhud [2]3 years ago

7 0

I believe it would be fifteen.

lions [1.4K]3 years ago

7 0

No it's 47.12, because if you look up the formula

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How is 0.87 rational

dmitriy555 [2]

A number is rational if it can be put into a fraction of two integers.
Since 0.87 can also be expressed as 87/100, it is rational

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3 years ago

Read 2 more answers

Use Stokes' Theorem to evaluate S curl F · dS. F(x, y, z) = 5y cos(z) i + ex sin(z) j + xey k, S is the hemisphere x2 + y2 + z2

Alenkinab [10]

By Stokes' theorem, the integral of the curl of $\vec F$ across $S$ is equal to the integral of $\vec F$ along the boundary of $S$ , call it $C$ . Parameterize $C$ by

$\vec r(t)=2\cos t\,\vec\imath+2\sin t\,\vec\jmath$

with $0\le t\le2\pi$ . So we have

$\displaystyle\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S=\int_C\vec F\cdot\mathrm d\vec r$

$=\displaystyle\int_0^{2\pi}(10\sin t\cos 0\,\vec\imath+e^{2\cos t}\sin0\,\vec\jmath+2\cos t\,e^{2\sin t}\,\vec k)\cdot(-2\sin t\,\vec\imath+2\cos t\,\vec\jmath)\,\mathrm dt$

$=\displaystyle\int_0^{2\pi}-20\sin^2t\,\mathrm dt$

$=\displaystyle-10\int_0^{2\pi}(1-\cos2t)\,\mathrm dt=\boxed{-20\pi}$

6 0

3 years ago

A pentagonal prism is cut be a plane perpendicular to the base . What is the shape of the cross section that is formed?

Allisa [31]

Answer:

Step-by-step explanation:

The cross section will have the same shape as the base of the pentagonal prism; the dimensions will be proportionally smaller.

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4 years ago

How many differet six letter arrangements can be made using the lettersof the world "Tatto"

Shtirlitz [24]

The only answer I could find was just tattoo. I even used an online unscrambler, that's it.

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3 years ago

What is the least multiple of 41 whose digits consist only of 1's

Likurg_2 [28]

Answer:

11111

Step-by-step explanation:

Multiple of a number is multiplication of the given number with any integer. These integers can be positives as well as negatives. Hence Multiples of any number can either be positive,negative or zero.

According to question,

We need to find a multiple of 41 which contains only 1's.

Hence, we will start by dividing a number greater than 41 which only contains 1's. i.e 111 and keep on adding 1 at unit place until the number is divisible by 41.

111÷41 gives 29 as remainder, hence is not completely divisible by 41.

∴ we add 1 to its unit place and use the same process for 1111.

1111÷41 is also not divisible by 41 as it leaves 4 as remainder. Hence we add 1 to its unit place and check for 11111.

11111÷41=271. Hence, 11111 is completely divisible by 41.

∵ 11111 is the least number which contains only 1's and is a multiple of 41

8 0

4 years ago