Please help I would really appreciate it.

I asked a question a while ago and nobody has answered it pls help

bogdanovich [222]

Answer:

bet but need barinlyist

Step-by-step explanation:

5 0

2 years ago

Pyramid A has a triangular base where each side measures 4 units and a volume of 36 cubic units. Pyramid B has the same height,

omeli [17]

Answer:

The volume of pyramid B is 81 cubic units

Step-by-step explanation:

Given

Pyramid A

$s = 4$ -- base sides

$V = 36$ -- Volume

Pyramid B

$s = 6$ --- base sides

Required

Determine the volume of pyramid B [Missing from the question]

From the question, we understand that both pyramids are equilateral triangular pyramids.

The volume is calculated as:

$V = \frac{1}{3} * B * h$

Where B represents the area of the base equilateral triangle, and it is calculated as:

$B = \frac{1}{2} * s^2 * sin(60)$

Where s represents the side lengths

First, we calculate the height of pyramid A

For Pyramid A, the base area is:

$B = \frac{1}{2} * s^2 * sin(60)$

$B = \frac{1}{2} * 4^2 * \frac{\sqrt 3}{2}$

$B = \frac{1}{2} * 16 * \frac{\sqrt 3}{2}$

$B = 4\sqrt 3$

The height is calculated from:

$V = \frac{1}{3} * B * h$

This gives:

$36 = \frac{1}{3} * 4\sqrt 3 * h$

Make h the subject

$h = \frac{3 * 36}{4\sqrt 3}$

$h = \frac{3 * 9}{\sqrt 3}$

$h = \frac{27}{\sqrt 3}$

To calculate the volume of pyramid B, we make use of:

$V = \frac{1}{3} * B * h$

Since the heights of both pyramids are the same, we can make use of:

$h = \frac{27}{\sqrt 3}$

The base area B, is then calculated as:

$B = \frac{1}{2} * s^2 * sin(60)$

Where

$s = 6$

So:

$B = \frac{1}{2} * 6^2 * sin(60)$

$B = \frac{1}{2} * 36 * \frac{\sqrt 3}{2}$

$B = 9\sqrt 3$

So:

$V = \frac{1}{3} * B * h$

Where

$B = 9\sqrt 3$ and $h = \frac{27}{\sqrt 3}$

$V = \frac{1}{3} * 9\sqrt 3 * \frac{27}{\sqrt 3}$

$V = \frac{1}{3} * 9 * 27$

$V = 81$

6 0

3 years ago

Read 2 more answers

At a local event, the ratio of hamburgers to hot dogs sold is 5:3. The number of hamburgers sold is 275. How many more hamburger

sveticcg [70]

*your
If the ratio is 5:3, then you divide 275 by 5, and then multiply it by 3. If you test the result, you'll find that the ratio fits.

4 0

3 years ago

On your wedding day you leave for the church 34 minutes before the ceremony is to begin, which should be plenty of time since th

Schach [20]

Answer:

Average speed is 37.35 mi/h

Step-by-step explanation:

given data

leave = 34 minutes before

church distance = 12.0 miles

average speed first 17 minutes = 5.0 mi/h

solution

so we find Total distance travel in first 17 minutes = speed × time

Total distance travel in first 17 minutes = 5 × $\frac{17}{60}$

Total distance travel in first 17 minutes = 1.416 mi

and

Distance Remaining = 12 - 1.416 = 10.584 mi

Time Remaining = 34 - 17 min = 17 min

so

remaining distance Average speed = $\frac{distance}{time}$

Average speed = $\frac{10.584}{\frac{17}{60}}$

Average speed is 37.35 mi/h

3 0

3 years ago

at the airport, parking lot a charges $3.00 to enter and $9.00 per day to park. parking lot b charges $10.00 to enter and $6.00

jenyasd209 [6]

Parking lot A equation- $3.00 + $9.00x
Parking lot B equation- $10.00 + $6.00x

Part A- For 3 days parking lot B would be the better option- $28

Part B- For 8 days parking lot B would also be the better option- $58

7 0

3 years ago

Read 2 more answers