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miskamm [114]
3 years ago
9

Please help me solve this problem ASAP!!

Mathematics
1 answer:
Komok [63]3 years ago
6 0

Answer:

y = 8x - 11

Step-by-step explanation:

Using the equation

 <u>y - y1 = m (x - x1)</u>

<u>y - 5 = 8 (x- 2)</u>

<u>y - 5 = 8x - 16</u>

Subtract the 5

y = 8x - 11

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Polynomials are algebraic expressions that consist of variables and coefficients. ... We can perform arithmetic operations such as addition, subtraction, multiplication and also positive integer exponents for polynomial expressions but not division by variable. An example of a polynomial with one variable is x2+x-12.

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3 years ago
Chris bought a new car for $11,000. If he
pshichka [43]

Answer

B

Step-by-step explanation:

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3 years ago
SOMEONE PLEASE HELP<br><br><br>and please explain it so I won't get confused further on ​
Pavlova-9 [17]
Number 5 is 110 degrees because it is the same angle as m<2.

Now, to find m<7
A straight line is 180 degrees. One side is already given. m<8+m<7 = 180 degrees. So 180-110=70.

Answers:
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5. m<8= 110 degrees

3 0
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

\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
PLEASE HELP !! <br><br> Will give the brainliest !!
Alika [10]

Answer:

A. They're not simalier because two pairs of corresponding angles in the two trapezoids are congurant

5 0
3 years ago
Read 2 more answers
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