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maksim [4K]
3 years ago
10

PLEASE HELP!! WILL MARK BRAINLIEST AND THANK YOU!!!

Mathematics
1 answer:
Semmy [17]3 years ago
4 0

Answer:

The answer is c because you would plug in -2 for x and -6 for y

Step-by-step explanation:

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Vikki [24]

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$90

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The total of all the items is $270. You are lucky that this is a multiple. Solve it like you were dividing 27 by 9, just add a 0. Good luck! - Justin <3

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Sara is hiring a babysitter who charges $20 per hour plus a $10 travel fee. What equation represents y, the amount of money the
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3 years ago
-9 more than a number results in -20
damaskus [11]

Answer:

-9+x=-20

Step-by-step explanation:

I am not so positive on this one, but if you are not doing inequalities, then this should be correct.

"More than" would insinuate that you are adding. In this class, you would have a number to replace that x, but since no number was given, you place the x in its place. x=-11, because when you "add" -9 with something to get -20, it would have to be -11. Negative numbers can be confusing because when you subtract a negative number by another negative, you end up with a negative. Like in this case. Normally, you would just put -11, so it would look like -9 - 11 = -20. But since this says "more than", unless you are doing inequalities, you add.

If you are doing inequalities, then your answer should be this:

-9 > x = -20

I hope this helps!

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4 0
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

\displaystyle=-2\int_{t=0}^{t=1}e^{(1-t)^2}(1-t)(t^2-2t+2)\,\mathrm dt

Complete the square in the quadratic term of the integrand: t^2-2t+2=(t-1)^2+1=(1-t)^2+1, then in the integral we substitute u=1-t:

\displaystyle=-2\int_{t=0}^{t=1}e^{(1-t)^2}(1-t)((1-t)^2+1)\,\mathrm dt

\displaystyle=-2\int_{u=0}^{u=1}e^{u^2}u(u^2+1)\,\mathrm du

Make another substitution of v=u^2:

\displaystyle=-\int_{v=0}^{v=1}e^v(v+1)\,\mathrm dv

Integrate by parts, taking

r=v+1\implies\mathrm dr=\mathrm dv

\mathrm ds=e^v\,\mathrm dv\implies s=e^v

\displaystyle=-e^v(v+1)\bigg|_{v=0}^{v=1}+\int_{v=0}^{v=1}e^v\,\mathrm dv

\displaystyle=-(2e-1)+(e-1)=-e

So, we have by the fundamental theorem of calculus that

\displaystyle\int_C\nabla f(x,y,z)\cdot\mathrm d\mathbf r=f(1,1,1)-f(0,0,0)

\implies-e=f(1,1,1)-2

\implies f(1,1,1)=2-e

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Lelechka [254]

Answer:

Step-by-step explanation:

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