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4 years ago

A rectangular pyramid and rectangular prism have the same base area and the same

Mathematics

2 answers:

denpristay [2]4 years ago

7 0

answer

The volume of the pyramid is 1/3 the volume of the prism.

Step-by-step explanation:

9966 [12]4 years ago

4 0

Answer:

B)The volume of the pyramid is 1/3 the volume of prism

Step-by-step explanation:

We are given that

Bases of rectangular pyramid and rectangular prism are same

Height of both rectangular pyramid and rectangular prism are same.

We know that

Volume of rectangular pyramid = $\frac{1}{3}bh$

Where b=Area of base

Volume of rectangular prism= $bh$

Where b=Area of base

Volume of rectangular pyramid= $\frac{1}{3}$ volume of rectangular prism

Hence, the volume of pyramid is 1/3 times the volume of volume of rectangular prism.

Option B is true.

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What is the product of –24 and 35?

beks73 [17]

Answer:

-840

Step-by-step explanation:

"Product" means to multiply, so you multiply -24 by 35. Since -24 has a negative, negative times a positive will equal a negative number.

(Negative times negative is positive.)

(Negative times positive is negative.)

(Positive times positive is obviously positive.)

Using these simple rules, we can do this:

Simply multiply -24 by 35, or you can multiply 24 by 35 which is 840, but DON'T forget to put a negative symbol!

Hope this helps! Let me know if there is anything else.

7 0

3 years ago

Read 2 more answers

When you rotate a set of parallel lines the new image must also be a set of parallel lines but they can be farther apart than th

Viktor [21]

I’m pretty sure it’s false.

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4 years ago

Wow the teacher gave me so much homework was worderin if u people could help me with a few problems..

vesna_86 [32]

1. a=-12
2. 120
3.-937

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3 years ago

Solve for a.<br><br> a+(−1.2)=−5.5<br><br> Enter your answer in the box.

Amanda [17]

Answer:

a = -4.3

Step-by-step explanation:

a+(−1.2)=−5.5

a - 1.2 = - 5.5

a = - 5.5 + 1.2

a = - 4.3

7 0

3 years ago

Solve (D^3-D^2-D-2)y=0

allsm [11]

(^3-^2+2)D=0

D^2+D^2+2D=0

2D^2+2D=0

a = 2; b = 2; c = 0;

Δ = b2-4ac

Δ = 22-4·2·0

Δ = 4

D1=−b−Δ√2aD2=−b+Δ√2a

Δ−−√=4√=2

D1=−b−Δ√2a=−(2)−22∗2=−44=−1

D2=−b+Δ√2a=−(2)+22∗2=04=0

8 0

3 years ago