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Marysya12 [62]
3 years ago
7

If you are looking for the height of a pyramid and the volume is 18,069,333.33333 and the length and width are both 440 what is

the height?
Mathematics
1 answer:
makkiz [27]3 years ago
5 0

Answer:

h=123,200

Step-by-step explanation:

First you must in the formula for the pyramid, which is

v=\frac{wlh}{3}

and as far as the question has gave us

V=18069333.33333

LW=440

1.) Plug those numbers into the the equation, which should be:

(18,069,333.33333)= \frac{(440)h}{3}

2.) Then isolate the "h" by dividing both sides by \frac{440}{3}, or multiplying its reciprocal, which is \frac{3}{440}

(\frac{3}{440}) 18,069,333.33333=\frac{440h}{3} (\frac{3}{440} )

3.) Simplify or round the numbers if you need to.

h= 123,200

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Answer:

b.

Step-by-step explanation:

hope this helps, could i get brainliest?

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3 years ago
A man standing on level ground is 1000 feet away from the base of a 350-foot-tall building. Find,
CaHeK987 [17]

If you sketch the man and the building on paper, you'll have a
right triangle.  The right angle is the point where the wall of
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from his feet to the top of the building is the hypotenuse.

We need to find the angle at his feet, between the hypotenuse
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Well, the side opposite the angle is the height of the building -- 350ft,
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The tangent of the angle is      (opposite) / (adjacent)

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To find the angle, use a book, a slide rule, a Curta, or a calculator
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3 0
3 years ago
Circle A is shown. Secant W Y intersects tangent Z Y at point Y outside of the circle. Secant W Y intersects circle A at point X
timama [110]

Circle A is missing, so i have attached it.

Answer:

∠XYZ = 35°

Step-by-step explanation:

We want to find the angle ∠XYZ in the image attached.

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From the image, arc WZ = 175° and arc XZ = 105°

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5 0
3 years ago
Read 2 more answers
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frez [133]
I hope this helps its step 2 gl
4 0
2 years ago
In a class of 50 students, everyone has either a pierced nose or a pierced ear. The professor asks everyone with a pierced nose
Tems11 [23]

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Step-by-step explanation:

Let E be the event of that student pierces ear and N be the event of that student pierces nose.

Given: n(E\cup N=50)

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For any two event A and B, we have

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50=46+7-n(E\cap N)\\\\\Rightarrow\ n(E\cap N)=53-50=3

Hence, 3 students have piercings both on their ears and their noses.

7 0
3 years ago
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