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3 years ago

If the altitude is √7, find the volume. 36√7 cu. units 18√7 cu. units 12√7 cu. units

Mathematics

1 answer:

rosijanka [135]3 years ago

8 0

Answer:

$Volume = 12\sqrt{7}$ cubic units

Step-by-step explanation:

Volume of the square pyramid is given as ⅓*area of base*height of pyramid.

Where,

Area of base = 6*6 = 36

Height of pyramid = altitude = $\sqrt{7}$ units

Plug the values into the formula to find volume.

$Volume = \frac{1}{3}*36*\sqrt{7}$

$Volume = 12*\sqrt{7}$

$Volume = 12\sqrt{7}$ cubic units

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PLEASE I NEED HELP ASAP

Annette [7]

The coordinates of the vertex that A maps to after Daniel's reflections are (3, 4) and the coordinates of the vertex that A maps to after Zachary's reflections are (3, 2)

<h3>How to determine the coordinates of the vertex that A maps to after the two reflections?</h3>

From the given figure, the coordinate of the vertex A is represented as:

A = (-5, 2)

<u>The coordinates of the vertex that A maps to after Daniel's reflections</u>

The rule of reflection across the line x = -1 is

(x, y) ⇒ (-x - 2, y)

So, we have:

A' = (5 - 2, 2)

Evaluate the difference

A' = (3, 2)

The rule of reflection across the line y = 2 is

(x, y) ⇒ (x, -y + 4)

So, we have:

A'' = (3, -2 + 4)

Evaluate the difference

A'' = (3, 4)

Hence, the coordinates of the vertex that A maps to after Daniel's reflections are (3, 4)

<u>The coordinates of the vertex that A maps to after Zachary's reflections</u>

The rule of reflection across the line y = 2 is

(x, y) ⇒ (x, -y + 4)

So, we have:

A' = (-5, -2 + 4)

Evaluate the difference

A' = (-5, 2)

The rule of reflection across the line x = -1 is

(x, y) ⇒ (-x - 2, y)

So, we have:

A'' = (5 - 2, 2)

Evaluate the difference

A'' = (3, 2)

Hence, the coordinates of the vertex that A maps to after Zachary's reflections are (3, 2)

Read more about reflection at:

brainly.com/question/4289712

#SPJ1

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2 years ago

Answer the following question CORRECTLY I will know if this is wrong. I WILL REPORT ANY INCORRECT ANSWERS!

vredina [299]

Answer is: B) An allowable amount of variation of a specified quantity, especially in the dimensions of a machine part.

Tolerance is how much something can take.

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3 years ago

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The square has a perimeter of 24 ft. What is the length of each side. Plsssss help

BigorU [14]

Each side is 6 feet.

let s be 1 side of the square

24 = 4s

6 = s

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3 years ago

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A training field is formed by joining a rectangle and two semicircles. The semicircles are attached on opposite sides of the rec

Anarel [89]

Step-by-step explanation:

There is figures in you question. Please check your question.

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3 years ago

A rectangular box with a volume of 272ft^3 is to be constructed with a square base and top. The cost per square foot for the bot

ASHA 777 [7]

Answer:

The dimensions of the box is 3 ft by 3 ft by 30.22 ft.

The length of one side of the base of the given box is 3 ft.

The height of the box is 30.22 ft.

Step-by-step explanation:

Given that, a rectangular box with volume of 272 cubic ft.

Assume height of the box be h and the length of one side of the square base of the box is x.

Area of the base is = $(x\times x)$

$=x^2$

The volume of the box is = area of the base × height

$=x^2h$

Therefore,

$x^2h=272$

$\Rightarrow h=\frac{272}{x^2}$

The cost per square foot for bottom is 20 cent.

The cost to construct of the bottom of the box is

=area of the bottom ×20

$=20x^2$ cents

The cost per square foot for top is 10 cent.

The cost to construct of the top of the box is

=area of the top ×10

$=10x^2$ cents

The cost per square foot for side is 1.5 cent.

The cost to construct of the sides of the box is

=area of the side ×1.5

$=4xh\times 1.5$ cents

$=6xh$ cents

Total cost = $(20x^2+10x^2+6xh)$

$=30x^2+6xh$

Let

C $=30x^2+6xh$

Putting the value of h

$C=30x^2+6x\times \frac{272}{x^2}$

$\Rightarrow C=30x^2+\frac{1632}{x}$

Differentiating with respect to x

$C'=60x-\frac{1632}{x^2}$

Again differentiating with respect to x

$C''=60+\frac{3264}{x^3}$

Now set C'=0

$60x-\frac{1632}{x^2}=0$

$\Rightarrow 60x=\frac{1632}{x^2}$

$\Rightarrow x^3=\frac{1632}{60}$

$\Rightarrow x\approx 3$

Now $C''|_{x=3}=60+\frac{3264}{3^3}>0$

Since at x=3 , C''>0. So at x=3, C has a minimum value.

The length of one side of the base of the box is 3 ft.

The height of the box is $=\frac{272}{3^2}$

=30.22 ft.

The dimensions of the box is 3 ft by 3 ft by 30.22 ft.

7 0

3 years ago