1. Write the first term of the arithmetic sequence whose first term is 7, and whose common difference is 9.

Mathematics

1 answer:

Elis [28]3 years ago

6 0

Answer:

first term: 7

find the 17th term : 102

fifteent term: 10

Step-by-step explanation:

hababa

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Yakvenalex [24]

Answer:

D. X-7

Step-by-step explanation:

The table tells you that when x=7, g(x) = 0. In order for g(7) to be zero, at least one factor must be zero when x=7. The only factor on the list that is zero when x=7 is (x-7).

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2 years ago

PLEASE HELPPP :(( lol

Ksivusya [100]

Answer:

the pic is block

Step-by-step explanation:

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2 years ago

Suppose that \nabla f(x,y,z) = 2xyze^{x^2}\mathbf{i} + ze^{x^2}\mathbf{j} + ye^{x^2}\mathbf{k}. if f(0,0,0) = 2, find f(1,1,1).

lesya [120]

The simplest path from (0, 0, 0) to (1, 1, 1) is a straight line, denoted $C$ , which we can parameterize by the vector-valued function,

$\mathbf r(t)=(1-t)(\mathbf i+\mathbf j+\mathbf k)$

for $0\le t\le1$ , which has differential

$\mathrm d\mathbf r=-(\mathbf i+\mathbf j+\mathbf k)\,\mathrm dt$

Then with $x(t)=y(t)=z(t)=1-t$ , we have

$\displaystyle\int_{\mathcal C}\nabla f(x,y,z)\cdot\mathrm d\mathbf r=\int_{t=0}^{t=1}\nabla f(x(t),y(t),z(t))\cdot\mathrm d\mathbf r$

$=\displaystyle\int_{t=0}^{t=1}\left(2(1-t)^3e^{(1-t)^2}\,\mathbf i+(1-t)e^{(1-t)^2}\,\mathbf j+(1-t)e^{(1-t)^2}\,\mathbf k\right)\cdot-(\mathbf i+\mathbf j+\mathbf k)\,\mathrm dt$