Find three consecutive odd integral such that the sum of seven times the smallest and twice the largest is -91
Answer:
1. C. Yes, because a sum of cubes can be factored
2a. false
2b. false
2c. true
2d. false (based on what is written in the equation; refer to step-by-step)
Step-by-step explanation:
1. Both 3 and 8 can be cubed, which is why x^3+8 can be factored (x+2)(x^2-2x+4)
2a. a^2-b^2 can be factored by the perfect square rule, so it should be (a-b)^2
2b. both terms are perfect squares, so you can factor, making it (a+b)(a-b)
2c. You can factor using the perfect square rule, making it (a+b)^2
2d. Most of what is in the equation is true, yet the correct solution would be (a-b)(a^2+ab+b^2)
What is the third term of the sequence defined by the recursive rule f(1)=3 f(n)=f(n-1)+4
Need f(2):
f(2)=f(2-1)+4
f(2)=f(1)+4
f(2)=(3)+4=7
FIND f(3):
f(3)=f(3-1)+4
f(3)=f(2)+4
f(3)=(7)+4
f(3)=11
Answer:
dont know
Step-by-step explanation:
sorry
