F(x) = 2^x; h(x) = x^3 + x + 8
Table
x f(x) = 2^x h(x) = x^3 + x + 8
0 2^0 = 1 0 + 0 + 8 = 8
1 2^1 = 2 1^3 + 1 + 8 = 10
2 2^2 = 4 2^3 + 2 + 8 = 8 + 2 + 8 = 18
3 2^3 = 8 3^3 + 3 + 8 = 27 + 3 + 8 = 38
4 2^2 = 16 4^3 + 4 + 8 = 76
10 2^10 = 1024 10^3 +10 + 8 = 1018
9 2^9 = 512 9^3 + 9 + 8 = 729 + 9 + 8 = 746
Answer: an approximate value of 10
Answer: 810
<u>Step-by-step explanation:</u>
There are 10 numbers to choose from. Since the 2nd digit and 3rd digit cannot repeat the number before it, they only have 9 numbers to choose from.
1st digit and 2nd digit and 3rd digit
10 choices x 9 choices x 9 choices = 810
Answer:
false
Step-by-step explanation:
Answer:
26,250
Step-by-step explanation:
25,000 + 5% = 26,250
:) please mark Brainliest
