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liubo4ka [24]
3 years ago
8

5 hundreds × 10 = __50__ hundreds = __B__ Fill in blank B.

Mathematics
1 answer:
zhenek [66]3 years ago
7 0

Answer:

Answer is 5000

Step-by-step explanation:

becaue when u multiple 5*100*10=5000

50*100=5000 so the answer is 5000

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the accountant charged $35 for the first hour of work and $23 for each hour he did after that he earned a totL or $127 how many
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He worked 5 hours in total: 35$ for the first hour, + 23 + 23 + 23 + 23 = 127$ for five hours of work. 
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If c is the curve given by \mathbf{r} \left( t \right = \left( 1 5 \sin t \right \mathbf{i} \left( 1 3 \sin^{2} t \right \mathbf
jonny [76]
With the curve C parameterized by

C:\mathbf r(t)=\underbrace{15\sin t}_{x(t)}\,\mathbf i+\underbrace{13\sin^2t}_{y(t)}\,\mathbf j+\underbrace{12\sin^3t}_{z(t)}\,\mathbf k

with 0\le t\le\dfrac\pi2, and given the vector field

\mathbf f(x,y,z)=x\,\mathbf i+y\,\mathbf j+z\,\mathbf k

the work done by \mathbf f on a particle moving on along C is given by the line integral

\displaystyle\int_C\mathbf f\cdot\mathrm d\mathbf r=\int\limits_{t=0}^{t=\pi/2}\mathbf f(x(t),y(t),z(t))\cdot\frac{\mathrm d\mathbf r(t)}{\mathrm dt}\,\mathrm dt

where

\mathrm d\mathbf r=(15\cos t\,\mathbf i+26\sin t\cos t\,\mathbf j+36\sin^2t\cos t\,\mathbf k)\,\mathrm dt

The integral is then

\displaystyle\int_0^{\pi/2}(15\sin t\,\mathbf i+13\sin^2t\,\mathbf j+12\sin^3t\,\mathbf k)\cdot(15\cos t\,\mathbf i+13\sin2t\,\mathbf j+18\sin t\sin2t\,\mathbf k)\,\mathrm dt
=\displaystyle\int_0^{\pi/2}(432\sin^5t\cos t+338\sin^3t\cos t+225\sin t\cos t
=269
6 0
3 years ago
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