Answer:
see below
Step-by-step explanation:
7,12,16,20,27,40,56,64
The square numbers are 16 = 4^2 and 64 = 8^2
The difference is 64 - 16 =48
The cube numbers are 27 = 3^3 and 64 = 4^3
64 - 27 =37
The rule of the <span>geometric sequence is
a ⇒⇒⇒ </span><span>the first term
r ⇒⇒⇒ </span><span>common ratio
Given: the fourth term </span>(a4) <span>= 121.5 and the common ratio = 3 and n = 4
∴</span><span>
∴ a = 121.5/3³ = 121.5/27 = 4.5
So, T</span><span>he formula for this sequence will be</span><span>
</span>
Answer:
Well, I don't think I can remember EVERY skill I was taught through the years but in Kindergarten we used Number Sense so we could accurately... also there's estimation and of course PEMDAS. I've also learned how to round up so that the solution is at least around those numbers if the skill is to be applied.
Step-by-step explanation:
I'm not sure if this'll help but go ahead and have it... just try and change it a bit so they don't get ya for plagirism
Answer:
We have the functions:
f(x) = IxI + 1
g(x) = 1/x^3.
Now, we know that the composite functions do not permute.
How we can prove this?
First, two composite functions are commutative if:
f(g(x)) = g(f(x))
Well, you could use brute force (just replace the values and see if the composite functions are commutative or not)
But i will use a more elegant way.
We can notice two things:
g(x) has a discontinuity at x = 0.
so:
f(g(x)) = I 1/x^3 I + 1
still has a discontinuty at x = 0, but:
g(f(x)) = 1/( IxI + 1)^3
here the denominator is IxI + 1, is never equal to zero.
So now we do not have a discontinuity.
Then the composite functions can not be commutative.
No because at least one segment length is not preserved.
rigid means everything stays the same, but orientation sometimes changes.
