Is this the answer u want?:
3a4+6a2+9ab2
Answer:
![\frac{3\sqrt{2} }{2}](https://tex.z-dn.net/?f=%5Cfrac%7B3%5Csqrt%7B2%7D%20%7D%7B2%7D)
Step-by-step explanation:
Given
+
← rationalise the denominator by multiplying by ![\frac{\sqrt{2} }{\sqrt{2} }](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%7B2%7D%20%7D%7B%5Csqrt%7B2%7D%20%7D)
=
+ (
×
)
=
+ ![\frac{2\sqrt{2} }{2}](https://tex.z-dn.net/?f=%5Cfrac%7B2%5Csqrt%7B2%7D%20%7D%7B2%7D)
= ![\frac{3\sqrt{2} }{2}](https://tex.z-dn.net/?f=%5Cfrac%7B3%5Csqrt%7B2%7D%20%7D%7B2%7D)
Answer:
My best guess would be B and C since those are the only numbers that aren't multiples of 5, and all the values of a(n) are.
A function is a relation that has one output for a given input.
For the first one, there is no one x value with two or more y values so it is a function.
The second example is also a function because a certain value can only have one cube root.
For problem number 3 input "-3" for every instance of x in h(x).
So, h(-3)=2(3^2)-1= 17
Answer:1/25
Step-by-step explanation: do 5 to the second power which will get all combinations then do 1 as the numerator over the denominator which is the number of combinations.