Let the 3-digit palindrome be 101a+10b. a+a+ab=8+b, so 2a+ab-b=8.
2a+b(a-1)=8; b=2(4-a)/(a-1). From this we know that a > 1. And we know 4-a > 0 so a < 4.
This limits a to 2 and 3 which make b=4 and 1.
The palindromes are 242 and 313.
Answer:
f(2) = 729
Step-by-step explanation:
f(x) = 3^(x + 4)
Let x =2
f(2) = 3^(2 + 4)
f(2) = 3^(6)
f(2) = 729
19) 5c+c=-6+2
6c=-4
c=-4/6
c=-2/3
20) -20-8=-2e
-28=-2e
28/2=e
14=e
21)p-2w=2l
(p-2w)/2=l
4500 = 4.5 * 10^3
57 = 5.7 * 10^1
730 = 7.3 * 10^2
0.007 = 7 * 10^-3
300.25 = 3.0025 * 10^2
56,325.2 = 5.63252 * 10^4
