The given sequence is
a₁ = 29
a₂ = 39
a₃ = 49
a₄ = 59
This sequence is an arithmetic sequence. Th first term is a₁ = 29, and the common difference is d= 10.
The n-th term is
The 33-rd termis
a₃₃ = 29 + (33 - 1)*10
= 29 + 320
= 349
Answer: a₃₃ = 349
Answer:
Option C.
The zeros, -24 and 24, can be found when 0= (x+24)(x-24).
Step-by-step explanation:
we have
we know that
To find the zeros, equate the equation to zero
square root both sides
therefore
1. We are given:
The only a which squared gives you 64 is 8. The perimeter of an 8 by 8 square will be 32. Half of that is 16. Now, let's see what we can do. We can set up:
Obviously, each side length has to be 4. So, the area of this square will be
16 units².
2. Let n equal her son's age. So, her age right now will be (S = her age):
1 year ago it was:
We have 2 equations, let's substitute. We can rewrite this as:
Solve for n:
We know the value of n, which is her son's age. So, her son is
1/15 of a year old or about 24 days old.
The ratio of juniors to seniors is 2<span> to 5</span>
Help please. ,,.,,.,.,.,.,.,.,
Umnica [9.8K]
It depends, are you trying to find area or perimeter?
