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2 years ago

What is the surface area of this rectangular pyramid?

Mathematics

1 answer:

Alexxx [7]2 years ago

7 0

Answer:

304m^2

Step-by-step explanation:

First find the surface area of the base by multiplying the length by the width.

(12m) (8m)= 96m^2

Second, find the surface area of the front and back triangles using the formula 1/2 (base) (height). Use the length for the base.

1/2 (12m) (10m)= 60m^2

Next, find the surface area of the side triangles using the formula 1/2 (base) (height). Use the width as the base.

1/2 (8m) (11m)= 44m^2

Last, add the surface area of each section. Make sure you add the area of each face.(we only solved for 1 of the front/ back triangles and 1 of the side triangles) To make it easier to understand I wrote out an equation to show how I added the surface areas.

base=a, front/ back triangles= b, side triangles=c

SA= a + 2b +2c or SA= a +b +b +c +c

Using one of the equations above solve for the total surface area.

SA= (96m^2) + (60m^2) +(60m^2) +(44m^2) +(44m^2)

SA= (96m^2) + 2(60m^2) +2(44m^2)

SA= (96m^2) +(120m^2) +(88m^2)

SA= 304m^2

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How many times smaller is 2 × 10^-10 than 4 × 10^-5

natta225 [31]

Answer:

200,000

Step-by-step explanation:

To find how many times smaller is smaller number than greater number, divide greater number by smaller number.

Greater number $=4\times 10^{-5}$

Smaller number $=2\times 10^{-10}$

Then

$\dfrac{\text{Greater number}}{\text{Smaller number}}\\ \\=\dfrac{4\times 10^{-5}}{2\times 10^{-10}}\\ \\=\dfrac{4}{2}\times \dfrac{10^{-5}}{10^{-10}}\\ \\=2\times \dfrac{1}{10^{-5}}\\ \\=2\times 10^5\\ \\=2\times 100,000\\ \\=200,000$

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3 years ago

the formula to convert a temperature C and degrees Celsius to a temperature F in degrees fahrenheit is 1.8C+32=F. Use this equat

Arturiano [62]

Answer:

Step-by-step explanation:

sorry this one isnt an easy one. but your answr should be 1.8

3 0

3 years ago

Write a Let" statement for the unknown, write an equation, and solve the equation. 11. Jane bought a number of records for $7 ea

Advocard [28]

Let n = number of records.

Each record costs $7, so n records cost 7n.

She then spent $3, so the total spent is 3n + 3.

She spent a total of $24, so 3n + 3 must equal 24. That gives us the following equation.

7n + 3 = 24

Subtract 3 from both sides.

7n = 21

Divide both sides by 7.

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Answer: She bought 3 records.

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3 years ago

A sphere and a cylinder have the same radius and height. The volume of the cylinder is . Amie found the volume of the sphere. He

dezoksy [38]

Amie's error while measuring the volume of a sphere is that Amie should have multiplied 54 by 2/3. Thus, the first option is the right choice.

In the question, we are given that a sphere and a cylinder have the same radius and height.

We assume the radius of the sphere to be r, and its height to be h.

Now, the height of a sphere is its diameter, which is twice the radius.

Thus, the height of the sphere, h = 2r

Given that the sphere and the cylinder have the same radius and height, the radius of the cylinder is r, and its height is 2r.

The volume of a sphere is given by the formula, V = (4/3)πr³, where V is its volume, and r is its radius.

Thus, the volume of the given sphere using the formula is (4/3)πr³.

The volume of a cylinder is given by the formula, V = πr²h, where V is its volume, r is its radius, and h is its height.

Thus, the volume of the given cylinder using the formula is πr²(2r) = 2πr³.

Now, to compare the two volumes we take their ratios, as

Volume of the sphere/Volume of the cylinder

= {(4/3)πr³}/{2πr³}

= 2/3.

Thus, the volume of the sphere/the volume of the cylinder = 2/3,

or, the volume of the sphere = (2/3)*the volume of the cylinder.

Given the volume of the cylinder to be 54 m³, Amie should have multiplied 54 by 2/3 instead of adding the two.

Thus, Amie's error while measuring the volume of a sphere is that Amie should have multiplied 54 by 2/3. Thus, the first option is the right choice.

Learn more about volumes at

brainly.com/question/12398192

#SPJ4

For the complete question, refer to the attachment.