Which statement is true about every rectangular pyramid?

Mathematics

2 answers:

fiasKO [112]3 years ago

7 0

Answer:

At least two of the lateral faces are congruent. This is the true statement.

Step-by-step explanation:

The correct statement is : At least two of the lateral faces are congruent.

A rectangular pyramid is a 3-D shape that has a rectangular base. Each corresponding side with respect to base, contains triangular faces. The triangular faces are called lateral faces and they meet at the apex. These faces are congruent to each other.

Lorico [155]3 years ago

4 0

The following statement is true: At least two of the lateral faces are congruent.

Because the words "at least" are used this means that at a minimum 2 of the faces are congruent. Because the base is in the shape of a triangle, this will definitely be true. The sides directly across from each other will be congruent and if the base is a square (which is also defined as a rectangle) than all 4 sides would be congruent.

You might be interested in

For his yearly vacation, Colton plans to spend $350 on train tickets, $200 on food, $150 on accommodations, and $50 on miscellan

scoundrel [369]

Answer: He would spend 750$ on the entire vacation.

Step-by-step explanation: Simple addition, 350+150 (Always do the two odd ones first) which equals 500, 500+200 equals 700, and 700+50 equals 750, which gives your answer of 750$'s for the whole vacation.

5 0

3 years ago

Read 2 more answers

WHAT IS THE REMAINDER WHEN <img src="https://tex.z-dn.net/?f=32%5E%7B37%5E%7B32%7D%20%7D" id="TexFormula1" title="32^{37^{32} }"

Feliz [49]

Recall Euler's theorem: if $\gcd(a,n) = 1$ , then

$a^{\phi(n)} \equiv 1 \pmod n$

where $\phi$ is Euler's totient function.

We have $\gcd(9,32) = 1$ - in fact, $\gcd(9,32^k)=1$ for any $k\in\Bbb N$ since $9=3^2$ and $32=2^5$ share no common divisors - as well as $\phi(9) = 6$ .

Now,

$37^{32} = (1 + 36)^{32} \\\\ ~~~~~~~~ = 1 + 36c_1 + 36^2c_2 + 36^3c_3+\cdots+36^{32}c_{32} \\\\ ~~~~~~~~ = 1 + 6 \left(6c_1 + 6^3c_2 + 6^5c_3 + \cdots + 6^{63}c_{32}\right) \\\\ \implies 32^{37^{32}} = 32^{1 + 6(\cdots)} = 32\cdot\left(32^{(\cdots)}\right)^6$