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3 years ago

Please help me with this I need it asap ! Find the surface area

Mathematics

1 answer:

sergij07 [2.7K]3 years ago

3 0

Answer:

Step-by-step explanation:

5 * 3 * 2 = 30

3 * 2 * 2 = 12

5 * 2 * 2 = 20

30 + 12 + 20 = 62

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Use properties of addition and subtraction to evaluate the expression <br> -42 - 29 - 28

Setler [38]

(-42) (-29) (-28)
-42 - 29 = -71
-71 - 28 = -99

-99 is your answer

hope this helps

7 0

3 years ago

Find the area of the shape shown below.

cricket20 [7]

Answer:

Step-by-step explanation:

3+9=12

12/2=6

6x3=18

3 0

3 years ago

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Math geometry show work Thanks

aev [14]

Problem 1, part (a)

<h3>Answer: False</h3>

For instance, 200 feet in real life can be reduced to scale down to say 2 inches on paper. So we have a reduction going on, and not an enlargement.

====================================================

Problem 1, part (b)

<h3>Answer: true</h3>

This is because a scale drawing involves similar polygons. This is true whenever any dilation is applied.

====================================================

Problem 2

I'm not sure how your teacher wanted you to answer this question. S/he didn't give you any numbers for the side lengths of the polygon. The angle measures are missing as well.

4 0

3 years ago

Find the curl of ~V<br> ~V<br> = sin(x) cos(y) tan(z) i + x^2y^2z^2 j + x^4y^4z^4 k

ch4aika [34]

Given

$\vec v = f(x,y,z)\,\vec\imath+g(x,y,z)\,\vec\jmath+h(x,y,z)\,\vec k \\\\ \vec v = \sin(x)\cos(y)\tan(z)\,\vec\imath + x^2y^2z^2\,\vec\jmath+x^4y^4z^4\,\vec k$

the curl of $\vec v$ is

$\displaystyle \nabla\times\vec v = \left(\frac{\partial h}{\partial y}-\frac{\partial g}{\partial z}\right)\,\vec\imath - \left(\frac{\partial h}{\partial x}-\frac{\partial f}{\partial z}\right)\,\vec\jmath + \left(\frac{\partial g}{\partial x}-\frac{\partial f}{\partial y}\right)\,\vec k$

$\nabla\times\vec v = \left(4x^4y^3z^4-2x^2y^2z\right)\,\vec\imath \\\\ - \left(4x^3y^4z^4-\sin(x)\cos(y)\sec^2(z)\right)\,\vec\jmath \\\\ + \left(2xy^2z^2+\sin(x)\sin(y)\tan(z)\right)\,\vec k$

$\nabla\times\vec v = \left(4x^4y^3z^4-2x^2y^2z\right)\,\vec\imath \\\\ + \left(\sin(x)\cos(y)\sec^2(z)-4x^3y^4z^4\right)\,\vec\jmath \\\\ + \left(2xy^2z^2+\sin(x)\sin(y)\tan(z)\right)\,\vec k$

7 0

3 years ago

Find the value of x for which ABCD must be a parallelogram.

Tems11 [23]

D. X = 8 , Y = 6
3x - 14 = x + 23x - x = 2 + 142x = 162x / 2 = 16/2x = 8
To check: 3x - 14 = x + 2 ; 3(8) - 14 = 8 + 2 ; 24 - 14 = 10 ; 10 = 10
4y - 7 = y + 114y - y = 11 + 73y = 183y / 3 = 18 / 3y = 6
To check: 4y - 7 = y + 11 ; 4(6) - 7 = 6 + 11 ; 24 - 7 = 17 ; 17 = 17
hopes it helps:)

8 0

4 years ago

Read 2 more answers