48 cubed as a product of its prime factors

1 answer:

6 0

48 cubed = 48*48*48

Find factors of 48:
12 and 4
12*4=48

Neither of those numbers are prime factors so we have to break both of then down further.

Find prime factors of 4:
2*2=4

Find factors and prime factors of 12:
12= 4*3
3 is a prime factors but 4 is not. Break down 4 again and get 2*2*3=12

So put that all together and get the prime factors of 48, they are:
2*2*2*2*3

The prime factor of 48 cubed is the same thing as the prime factor of 48 three times

The prime factors are:
2*2*2*2*2*2*2*2*2*2*2*2*3*3*3
(12 2's and 3 3's)

You might be interested in

In Missy's sports card collection, 3/4 of the cards are baseball cards. In Frank's collection, 8/12 are baseball cards. Frank sa

pochemuha

Answer: No he is incorrect because Missy's collection has more.

Step-by-step explanation:

3 0

3 years ago

The two points (-2,4) and (4,2) are the endpoints of the diameter of a circle. What is the equation of this circle in standard f

stepladder [879]

$Center\ of\ the\ circle\ is\ the\ midpoint\ AB\ where\ A(-2;\ 4)\ \&\ B(4;\ 2)\\\\center:O\left(\frac{-2+4}{2};\ \frac{4+2}{2}\right)\to O(1;\ 3)\\\\Radius\ of\ the\ circle\ is\ distance\ O\ to\ A:\\\\r=|OA|=\sqrt{(-2-1)^2+(4-3)^2}=\sqrt{9+1}=\sqrt{10}\\\\Equation\ of\ a\ circle:\\\\(x-a)^2+(y-b)^2=r^2\ whete\ center\ is\ (a;\ b)\ and\ radius\ is\ r.\\\\Equation\ of\ this\ circle:\\\\(x-1)^2+(y-3)^2=(\sqrt{10})^2\\\\(x-1)^2+(y-3)^2=10$