It's a trick question. There are an infinite number of mixed numbers between 3 and 4 that can multiply to equal 12 (for example, 3 and 3/7 times 3 and 1/2), but there are no mixed numbers between 3 and 4 that can multiply to equal 9. 3 times 3 is not between them but is 3, but that quantity is excluded because 3<x<4. Anything even a small bit above the number 3 would have to be multiplied by 2 and some fraction, which would not be between 3 and 4.
to answer on cube root we need to use the prime factorisation.
which means using only prime number and dividing the number by suitable prime number.
after find prime factorisation we know that cube root is 3th power of number.
so if we have same three number we need to take it as one because it is a cube root .
for example![\sqrt[3]{5*5*5} =\sqrt[3]{5^3} =5.](https://tex.z-dn.net/?f=%20%5Csqrt%5B3%5D%7B5%2A5%2A5%7D%20%3D%5Csqrt%5B3%5D%7B5%5E3%7D%20%3D5.%20)
Answer:
36
Step-by-step explanation:
Multiply the base thats 9 with the length that's 4
9*4 = 36
Answer:
f(x) = 2(x+3)+2
Step-by-step explanation:
