The first term, a, is 2. The common ratio, r, is 4. Thus,

a_(n+1) = 2(4)^(n).

Check: What's the first term? Let n=1. Then we get 2(4)^1, or 8. Is that correct? No.

Try this instead:

a_(n) = a_0*4^(n-1). Is this correct? Seeking the first term (n=1), does this formula produce 2? 2*4^0 = 2*1 = 2. YES.

The desired explicit formula is a_(n) = a_0*4^(n-1), where n begins at 1.

3 0

3 years ago

Please someone help me out. I beggggg

bezimeni [28]

Answer:

1. 144 2. 16 3. 1 4. 3x-6

Step-by-step explanation:

So think of this as a function in a function. So you work from the inside to the outside. So for problem 1, we start with f(4)) [you read it "f of 4"] so what is the solution when x = 4, since f(x) means the function of x so f(4) means 'the function of 4' inside f(x).

Since f(x) = 3x then f(4) = 3(4) [notice how you substitute the 4 everywhere you see a letter x]

so f(4) = 12, now you work the next part h(f(4)) since f(4)=12 then h(12)

So take the h(x) function which is h(x) = $x^{2}$ then h(12) = $12^{2}$ so h(12) = 144

4 0

2 years ago