$\bf \begin{array}{clclll}
3^{1001}&\qquad &4^{1002}\\
\uparrow &&\uparrow \\
a&&b
\end{array}\\\\
-----------------------------\\\\
(a+b)^2-(a-b)^2\implies (a^2+2ab+b^2)-(a^2-2ab+b^2)
\\\\\\
a^2+2ab+b^2-a^2+2ab-b^2\implies 2ab+2ab\implies 4ab
\\\\\\
now\qquad 4(3^{1001})(4^{1002})=k12^{1001}\qquad 
\begin{cases}
12^{1001}\\
(4\cdot 3)^{1001}\\
4^{1001}\cdot 3^{1001}
\end{cases}$

$\bf \\\\\\
4(3^{1001})(4^{1002})=k(4^{1001}\cdot 3^{1001})\implies \cfrac{4(3^{1001})(4^{1002})}{4^{1001}\cdot 3^{1001}}=k
\\\\\\
4\cdot \cfrac{3^{1001}}{3^{1001}}\cdot \cfrac{4^{1002}}{4^{1001}}=k\implies 4\cdot 4^{1002}4^{-1001}=k\impliedby 
\begin{array}{llll}
\textit{same base}\\
\textit{add the}\\
exponents
\end{array}
\\\\\\
4\cdot 4^{1002-1001}=k\implies 4\cdot 4^1=k\implies 16=k$

6 0

3 years ago

Please help with math

bezimeni [28]

Answer:

First, third, and fourth one would be the answers.

4 0

3 years ago