A24 = 141,
using equation,
an = a1 + (n-1)d,
a8 = a + 7d = 45
a16 = a + 15d = 93
solve simultaneously for values a and d, where a = 3, and d = 6.
therefore, inserting values into a24 eqn, where
a24 = 3 + (24-1)(6) = 141.
Number: n
square of number: n^2
sum of n and n^2 is n+n^2=20
Rewriting this equation, we get n^2+n-20=0 = (n+5)(n-4) = 0
Then n+5=0 and n-4=0, so n = -5 and n = 4.
You must check both results. It could happen that both are correct, or that only one is correct.
Answer:
B
Step-by-step explanation: 6 21 9 45
49 × 17 + 49 × 3
49 × (17 + 3)
49 × 20
It’s times by 5 just graph the points on the graph then count how much the rise and run is and then you have your slop
