3 years ago

Brandon is thinking of a number that is divisible by 6 and 8 . What is the smallest number that Brandon could think of

Mathematics

1 answer:

a_sh-v [17]3 years ago

4 0

since

6 = 2 * 3 <--- two factors

8 = 2 * 2 * 2 <--- three factors

and since the "2" is there on each, then we can just use that instance of it once.

so instead of using 2 * 3 * 2 * 2 * 2, we'll use 2 * 3 * 2 * 2, and that is our LCD.

2 * 3 * 2 * 2 = 24.

You might be interested in

I need help I can’t figure it out

dusya [7]

Answer:

$\large\boxed{\dfrac{41p}{70q^4(r-9)^2}}$

Step-by-step explanation:

$\dfrac{2p^4}{5q^5(r-9)^3}\cdot\dfrac{41q(r-9)}{28p^3}\\\\=\dfrac{2\!\!\!\!\diagup^1\cdot41}{5\cdot28\!\!\!\!\!\diagup_{14}}\cdot\dfrac{q\!\!\!\!\diagup}{q^{5\!\!\!\!\diagup^4}}\cdot\dfrac{(r\!\!\!\!\!\!{--}-9)\!\!\!\!\!\!{--}}{(r-9)^{3\!\!\!\!\diagup^2}}\cdot\dfrac{p^{4\!\!\!\!\diagup}}{p^3\!\!\!\!\!\!\diagup}\\\\=\dfrac{41}{70}\cdot\dfrac{1}{q^4}\cdot\dfrac{1}{(r-9)^2}\cdot\dfrac{p}{1}\\\\=\dfrac{41p}{70q^4(r-9)^2}$