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2 years ago

Pls help me on this asap

Mathematics

1 answer:

just olya [345]2 years ago

8 0

Answer:

I'm pretty sure it would be 0.83 mi/h

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What is m Any answers?

Likurg_2 [28]

Answer:

B. 68°

Step-by-step explanation:

Since, opposite sides of the quadrilateral MNPQ are parallel.

Therefore, it is a parallelogram.

Measures of the opposite angles of a parallelogram are equal.

So,

(6x - 2)° = (4x + 36)°

6x - 2 = 4x + 36

6x - 4x = 36 + 2

2x = 38

x = 38/2

x = 19

$m\angle M = (6x - 2)\degree \\ \\ m\angle M = (6 \times 19- 2)\degree \\ \\ m\angle M = (114- 2)\degree \\ \\ m\angle M =112\degree \\ \\ \because m\angle M + m\angle N = 180\degree \\ (adjacent \: angles) \\ \\ \therefore m\angle N = 180\degree - m\angle M \\ \\ \therefore m\angle N = 180\degree - 112 \degree \\ \\ \therefore m \angle N = 68\degree$

8 0

2 years ago

Does anybody know this?

vagabundo [1.1K]

Answer:

see attached

Step-by-step explanation:

Reflection in the line y=1 is accomplished by the transformation ...

(x, y) ⇒ (x, 2 -y)

Each point in the image is as far below the line as the original point is above the line.

6 0

2 years ago

3(x - 5) = 30<br> I dont know how to solve this

Anastaziya [24]

Answer:

x=15

Step-by-step explanation:

3(x-5)=30

First, you would distribute the three to the x and the negative five.

3x-15=30

Then you would add 15 to the 15 to cancel it and to the 30 to make it even.

3x-15=30

+15=+15=

3x=45

Finally, you would divide 3 into both sides, cancelling out the three and fiving you,

x=15

Hope this helped!

3 0

2 years ago

Read 2 more answers

Evaluate c (y + 7 sin(x)) dx + (z2 + 9 cos(y)) dy + x3 dz where c is the curve r(t) = sin(t), cos(t), sin(2t) , 0 ≤ t ≤ 2π. (hin

saw5 [17]

Treat $\mathcal C$ as the boundary of the region $\mathcal S$ , where $\mathcal S$ is the part of the surface $z=2xy$ bounded by $\mathcal C$ . We write

$\displaystyle\int_{\mathcal C}(y+7\sin x)\,\mathrm dx+(z^2+9\cos y)\,\mathrm dy+x^3\,\mathrm dz=\int_{\mathcal C}\mathbf f\cdot\mathrm d\mathbf r$

with $\mathbf f=(y+7\sin x,z^2+9\cos y,x^3)$ .

By Stoke's theorem, the line integral is equivalent to the surface integral over $\mathcal S$ of the curl of $\mathbf f$ . We have

$\nabla\times\mathbf f=(-2z,-3x^2,-1)$

so the line integral is equivalent to

$\displaystyle\iint_{\mathcal S}\nabla\times\mathbf f\cdot\mathrm d\mathbf S$
$=\displaystyle\iint_{\mathcal S}\nabla\times\mathbf f\cdot\left(\dfrac{\partial\mathbf s}{\partial u}\times\dfrac{\partial\mathbf s}{\partial v}\right)\,\mathrm du\,\mathrm dv$

where $\mathbf s(u,v)$ is a vector-valued function that parameterizes $\mathcal S$ . In this case, we can take

$\mathbf s(u,v)=(u\cos v,u\sin v,2u^2\cos v\sin v)=(u\cos v,u\sin v,u^2\sin2v)$

with $0\le u\le1$ and $0\le v\le2\pi$ . Then

$\mathrm d\mathbf S=\left(\dfrac{\partial\mathbf s}{\partial u}\times\dfrac{\partial\mathbf s}{\partial v}\right)\,\mathrm du\,\mathrm dv=(2u^2\cos v,2u^2\sin v,-u)\,\mathrm du\,\mathrm dv$

and the integral becomes

$\displaystyle\iint_{\mathcal S}(-2u^2\sin2v,-3u^2\cos^2v,-1)\cdot(2u^2\cos v,2u^2\sin v,-u)\,\mathrm du\,\mathrm dv$
$=\displaystyle\int_{v=0}^{v=2\pi}\int_{u=0}^{u=1}u-6u^4\sin^3v-4u^4\cos v\sin2v\,\mathrm du\,\mathrm dv=\pi$ <span />

4 0

2 years ago

What is twice 230? Write using exponents. Explain your reasoning.

Zielflug [23.3K]

2 x 230 = 231
I’m not 100% sure if this is correct?? Hope it helps though

6 0

3 years ago