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3 years ago

What are the measures of ∠1, ∠2, and ∠3? Enter your answers in the boxes.

Mathematics

1 answer:

GarryVolchara [31]3 years ago

3 0

Answer: m<1 is 60, m<2 is 60, and m<3 is 120

Step-by-step explanation:

M<1 is 60 and m<2 is also 60 because the interior triangle angles have to equal 180 so M<1 + m<2 + 60 = 180 so m<1 and m<2 are both sixty.

m<3 is 120 because m<2 and m<3 are supplementary so if a supplementary angle is equal to 180 then m<2 + m<3 = 180 and m<2 is 60 so m<3 is 120.

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PLEASE HELP!!!! WILL MARK BRAINLIEST!!!!

irga5000 [103]

Answer:

None of these.

Step-by-step explanation:

Let's assume we are trying to figure out if (x-6) is a factor. We got the quotient (x^2+6) and the remainder 13 according to the problem. So we know (x-6) is not a factor because the remainder wasn't zero.

Let's assume we are trying to figure out if (x^2+6) is a factor. The quotient is (x-6) and the remainder is 13 according to the problem. So we know (x^2+6) is not a factor because the remainder wasn't zero.

In order for 13 to be a factor of P, all the terms of P must be divisible by 13. That just means you can reduce it to a form that is not a fraction.

If we look at the first term x^3 and we divide it by 13 we get $\frac{x^3}{13}$ we cannot reduce it so it is not a fraction so 13 is not a factor of P

None of these is the right option.

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3 years ago

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Rectangle JKLM is similar to rectangle WXYZ . What is the scale factor of a dilation from JKLM to WXYZ ? Enter your answer in th

Ierofanga [76]

The answer is 1.5 because I just took the test.

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3 years ago

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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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3 years ago

The stock market dropped 220 points on Monday. It lost another 48 points on Tuesday, and on Wednesday, it gained 96 points. What

MissTica

48 points gained in those 3 days.

Subtract 220 with 48, then add 96. Last step would be subtracting 220 with the amount they have now in those 3 days to find how much points the stock market gained in those 3 days.

Subtract 220-48.It‘ll be 172.

Add 172 with 96. 172+96=268.

Subtract 268 with how much points the stock market had 3 days ago. 268-220.

Sot eh stock market gained 48 points in 3 days.

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3 years ago

Use the equation p = 8,31 where p = pressure and

djyliett [7]

Answer:

1 The pressure approaches infinity.

2 The function is undefined for V = 0.

3 There is an asymptote at V = 0.

tbh google =)

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3 years ago