Baking time in minutes
![]() ![]() |
![]() |
Deal with the formula in the parentheses first:
3 Δ 1 = 31 + 1 = 3 + 1 = 4
−2 Δ (3 Δ 1) = −2 Δ 4
−2 Δ 4 = (−2)4 + 4 = 16 + 4 = 20
No numbers are specified, so you should choose values for the original distance and the original rate. Good numbers to pick for the distance are multiples of 2, because the rate is decreased by
. Good numbers to pick for the rate are multiples of 3, because the rate is increased by
:
Old | New | |
---|---|---|
Distance | 12 | 6 |
Rate | 3 | 4 |
Time = Distance/Rate | 12/3 = 4 | 6/4 = 1.5 |
![]() |
![]() |
S11 = 2 × (11) − 5 = 22 − 5 = 17
B9 = (−1)(9) × 9 + 3 = −9 + 3 = −6
You know the value of A4, therefore, you can write the definition for A4 to solve for A3, write the definition for A3 to solve for A2, and so on:
There are 114 − 3 = 111 “jumps” of 4 between the 3rd and the 114th terms. Because 111 × 4 = 444, there is an increase of 444 from the 3rd term to the 114th term: 13 + 444 = 457.
The units digits of the powers of 7 follow a repeating pattern: 7, 49, 343, 2401, 16807, etc. Pattern = {7, 9, 3, 1}. There are 8 repeats of the pattern from A1 to A32, inclusive. The pattern begins again on A33, so A33 has the same units digit as A1, which is 7. The units digit of 733 is 7, and 7 − 1 = 6.
Simply plug in (−1) for each occurrence of x in the function definition and evaluate:
Plug in 3 for u in the definition of t(u), set it equal to 37, and solve for a:
t(u) = au2 − 3u + 1 → t(3) = a(3)2 − 3(3) + 1 = 37
9a − 9 + 1 = 37
9a = 45
a = 5
First, find the output value of the inner function:
Then, find g(10): 102 = 100.
Simply plug in
for y in g(y), and simplify the expression: