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Bingel [31]
3 years ago
13

8675 divested by .50 solved

Mathematics
2 answers:
vladimir2022 [97]3 years ago
7 0
It is 173.5 because if youl mulitipy 2 times 173.5 it equal to 8,675
Gnoma [55]3 years ago
6 0
8675 ÷ .5 = 17350
(you could also multiply by 2)
to check your answer:
17350 x .5 = 8675
(you could also divide by 2 #samething)
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HELP PLEASE <br> solve the right triangle
aleksandrvk [35]

Answer:

Part 1) FG=4.4\ units

Part 2) EF=3.9\ units

Part 3) m\angle G=63^o

Step-by-step explanation:

step 1

Find the measure of length side FG

In the right triangle EFG

we know that

sin(27^o)=\frac{GE}{FG} ----> by SOH (opposite side divided by the hypotenuse)

substitute the given values

sin(27^o)=\frac{2}{FG}

FG=\frac{2}{sin(27^o)}=4.4\ units

step 2

Find the measure of length side EF

In the right triangle EFG

we know that

cos(27^o)=\frac{EF}{FG} ----> by CAH (adjacent side divided by the hypotenuse)

substitute the given values

cos(27^o)=\frac{EF}{4.4}

EF=cos(27^o)(4.4)=3.9\ units

step 3

Find the measure of angle G

we know that

m\angle G+27^o=90^o ---> by complementary angles in a right triangle

m\angle G=90^o-27^o=63^o

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3 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

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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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Step-by-step explanation:

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I'm not sure if I did this right.

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