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3 years ago

Which three-dimensional figure has half the number of vertices as the rectangular prism?

Mathematics

2 answers:

GuDViN [60]3 years ago

8 0

1. The rectangular prism is a right prism with rectangular bases (check the picture)

2. It has 8 vertices, so we are looking for 3d figures with 4 vertices.

3. A face can have at least 3 vertices, and the 4th must be outside of the plane determined by the first 3 vertices. This makes 4 vertices in total.

4. The 3d figure described in step 3 is a pyramid.

svlad2 [7]3 years ago

4 0

The correct answer is:

D. Triangular pyramid.

The three-dimensional figure that has half the number of vertices as the rectangular prism is a triangular pyramid.

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Question 7 of 12 64 = O A. 9 OB. 7 C. 8 O D. 16 Pleaseee

sasho [114]

Answer:7

Step-by-step explanation:

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3 years ago

Read 2 more answers

Question 1,2, and 3 how do i factor those? Can you show the work and explain how?

DiKsa [7]

1: $3n^{2}+9n+6$

notice that each part is divisible by 3

$3n^{2}$ ÷ 3 = $n^{2}$

9n ÷ 3 = 3n

6 ÷ 3 = 2

so it becomes $3(n^{2} +3n+2)$

3n can be rewritten as 2n+n

-you want to rewrite it into two numbers that multiply to the number that's alone (in this case 2)

which would get you

$3(n^{2} +2n+n+2)$

Now that it's rewritten, you can factor out n + 2 from the equation.

the answer is

3(n+2)(n+1)

And you can check that by multiplying (n+2)(n+1) which is $n^{2} +2n+n+2$ and then each of those by 3, which is $3n^{2} +6n+3n+6$ or $3n^{2}+9n+6$ , our origional equation

2: $28+x^{2} -11x$

So I rewrote this as $x^{2} -11x+28$ (it's the same thing, just reordered using the commutative property)

now -11x can be rewritten as -4x-7x

(remember, the two numbers should multiply to equal 28, which is our constant.)

$x^{2} -4x-7x+28$

now we can factor out x from the first expression and -7 from the second

$x(x-4)-7(x-4)$

and lastly you factor out x-4,

which would give you

(x-4)(x-7)

Make sure to check your work and make sure it multiplies to $x^{2} -11x+28$

3: $9x^{2} -12x+4$

The first thing I notice when looking at this problem is that both 9 and 4 are perfect squares. Not only that, but they are the squares of 2 and 3, which are products of -12

So if you rewrite 9 as $3^{2}$ and 4 as $2^{2}$ , the equation becomes