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3 years ago

Ned created a video that was originally 40 minutes

Mathematics

2 answers:

iren2701 [21]3 years ago

8 0

Answer:

36.8 minutes long

Step-by-step explanation:

8 percent is 3.2

40 - 3.2 = 36.8

Zanzabum3 years ago

3 0

Answer:

36.8 minutes or 36 minutes 48 seconds

Step-by-step explanation:

8 percent of 40 is 3.2

(.08 times 40)

40 minus 3.2 is 36.8 :)

(.8 times 60 is 48)

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Cooling towers are used to remove or expel heat from a process. A cooling tower's walls are modeled by x squared over 324 minus

olchik [2.2K]

The width of the cooling tower at the base of the structure will be A. 36 meters.

<h3>How to calculate the width?</h3>

From the information given, the towers walls are modeled by x²/324 - (y² - 90)²/1600.

Therefore, the width of the cooling tower at the base of the structure will be:

= 2 × ✓324

= 2 × 18

= 36

In conclusion, the width is 36 meters.

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2 years ago

Which of the following expression are equivalent to -45+12?

lora16 [44]

Answer:

Step-by-step explanation:

Please share the possible answer choices next time. Thanks.

-45 + 12 is equivalent to 12 - 45.

-45 + 12 comes out to -33.

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3 years ago

What is the domain of the function below? (0,2 (3,1) (5,2) (8,4)

KiRa [710]

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3 years ago

Read 2 more answers

-1/4×3/5×-2/5 what dose this Equal

Makovka662 [10]

Answer:

0.06

Step-by-step explanation:

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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

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

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

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

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3 years ago