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2 years ago

PLEASE HELP ME!!!!!!!!!!!

Mathematics

2 answers:

erma4kov [3.2K]2 years ago

8 0

Answer: to find f(-4)

umka21 [38]2 years ago

4 0

Answer:

The answer is -4

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Please help me answer this I don't understand, I would really appreciate how you solved it as well.

abruzzese [7]

Answer: Parallel

Step-by-step explanation:

Set your -2x+10y=5 equation equal to y

-2x+10y=5

-10y -10y

-2x=5-10y

-5 -5

-2x-5+-10y

-10 -10 -10

-1/5x-1/2=y

Now you can put your equations into your graphing calculator and examine the lines made

(if you dont have one search up online graphing calculator on google or your app store)

5 0

3 years ago

Solve for d. d+ 4 = 16.21 d=

lutik1710 [3]

Answer:

d = 12.21

Step-by-step explanation:

1) Subtract 4 from both sides.

d = 16.21 - 4

2) Simplify 16.21 - 4 to 12.21.

d = 12.21

Check the answer!

⇒ d + 4 = 16.21

1. Let d = 12.21.

⇒ 12.21 + 4 = 16.21

2. Simplify 12.21 + 4 to 16.21

16.21 = 16.21

Done!

Thanks,

Eddie

8 0

11 months 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

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

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

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

3 0

3 years ago

Which shows two triangles that are congruent by AAS? A, B, C, or D

Iteru [2.4K]

The Answer is A. Hope I helped!

3 0

3 years ago

Read 2 more answers

F(x)=x^2 -5x+1 f(x+h)-f(x)/h =

Sergeu [11.5K]

The answer is

F = x - 5 + 1/5

8 0

3 years ago