What is the coefficient of the second term of the trinomial?

Mathematics

2 answers:

Schach [20]3 years ago

6 0

This is easy since you are already given the binomials. Just multiply them together.
(3x + 3)(3x + 3)
9x^2 + 9x + 9x + 9
9x^2 + 18x + 9.
The coefficient of term b is 18, so B = 18

spayn [35]3 years ago

3 0

Answer:

Step-by-step explanation:

You might be interested in

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,

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