Which three-dimensional shape has 5 vertices?

Mathematics

1 answer:

Levart [38]3 years ago

5 0

Answer:

C. Triangular pyramid

Four base vertices and the upper peak vertice

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The equation $a^7xy-a^6y-a^5x=a^4(b^4-1)$ is equivalent to the equation $(a^mx-a^n)(a^py-a^2)=a^4b^4$ for some integers $m$, $n$

Ksivusya [100]

Let simplify the equations $a^7xy-a^6y-a^5x=a^4(b^4-1)$ and $(a^mx-a^n)(a^py-a^2)=a^4b^4$ :

1) $a^7xy-a^6y-a^5x+a^4=a^4b^4$

and

2) $a^{m+p}xy-a^{n+p}y-a^{m+2}x+a^{n+2}=a^4b^4$ .

Equate the coefficients:

 $xy: m+p=7 \\ y: n+p=6 \\ x: m+2=5 \\ 1: n+2=4$ .
Then $n=2 \\ p=4 \\ m=3$ and mnp=24.

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3 years ago

How can you find the radius of this shape?

V125BC [204]

Answer:

Umm i think yu but it... I would hae to physcally be with you to help you.

Step-by-step explanation:

Sorry. Its hard to explain...

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3 years ago

Simplify the polynomial expressions by combining like terms and then multiply the resulting binomial expressions to find their p

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