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IRISSAK [1]
3 years ago
11

Send help!!!!!!!!!!!!!!

Mathematics
2 answers:
Veronika [31]3 years ago
7 0
Hello !

(17^{ \frac{1}{7}})^{7}  = \\  \\   17^{ \frac{1}{7}*7 } =   \\  \\ = 17^{1} \\  \\  = 17


From the formula : (a^{m})^{n} =a^{m*n}

I hope it helped:)

Diano4ka-milaya [45]3 years ago
5 0
Powers of power multiply:

(17^{1/7})^7 = 17^{(\frac 1 7 \cdot 7)} = 17^1 = 17

Answer: 17


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a) Estimate the volume of the solid that lies below the surface z = 7x + 5y2 and above the rectangle R = [0, 2]⨯[0, 4]. Use a Ri
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In the x direction we consider the m=2 subintervals [0, 1] and [1, 2] (each with length 1), while in the y direction we consider the n=2 subintervals [0, 2] and [2, 4] (with length 2). Then the lower right corners of the cells in the partition of R are (1, 0), (2, 0), (1, 2), (2, 2).

Let f(x,y)=7x+5y^2. The volume of the solid is approximately

\displaystyle\iint_Rf(x,y)\,\mathrm dx\,\mathrm dy\approx f(1,0)\cdot1\cdot2+f(2,0)\cdot1\cdot2+f(1,2)\cdot1\cdot2+f(2,2)\cdot1\cdot2=\boxed{164}

###

More generally, the lower-right-corner Riemann sum over m=\mu and n=\nu subintervals would be

\displaystyle\sum_{m=1}^\mu\sum_{n=1}^\nu\left(7\frac{2m}\mu+5\left(\frac{4n-4}\nu\right)^2\right)\frac{2-0}\mu\frac{4-0}\nu=\frac83\left(101+\frac{21}\mu+\frac{40}{\nu^2}-\frac{120}\nu\right)

Then taking the limits as \mu\to\infty and \nu\to\infty leaves us with an exact volume of \dfrac{808}3.

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The answer to this question would be marked as true
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