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jekas [21]
3 years ago
15

I will never get the hang of these questions help

Mathematics
1 answer:
jeyben [28]3 years ago
4 0
First term is -7, so a_1 = -7

To get the next term, we add on 4. We can see this if we subtract like so
d = (2nd term) - (1st term) = (-3) - (-7) = -3+7 = 4

So d = 4 is the common difference. 

Apply a_1 = -7 and d = 4 to get...
a_n = a_1 + d*(n-1)
a_n = -7 + 4*(n-1)
a_n = -7 + (n-1)*4

Answer: Choice A
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Nancy and Kevin collect coins. Nancy has x coins. Kevin has 9 coins fewer than five times the number of coins Nancy has. Write a
Alexus [3.1K]

Answer:

x + (5x-9)

Step-by-step explanation:

x stands for the number of coins Nancy has

Kevin has 9 less than 5 times what Nancy has, so his part of the expression is 5x-9

To find the total number of coins they have, you have to add Nancy's number of coins x plus Kevin's number of coins 5x-9

x + (5x-9)

But you have to put parentheses around Kevin's 5x-9 so that part of the expression is calculated first, as per order of operations

3 0
2 years ago
Jogging burns 375 calories in 45 minutes. Determine how many calories burned per hour?
Sveta_85 [38]
Answer: 500 calories

Set them as proportions
375/46=x/60
Multiply 60 to each side to eliminate fraction
Then you’ll get x=500
3 0
2 years ago
Here are 100 seats on an airplane, and each of 100 passengers has a ticket for a different seat. The passengers line up to board
Kryger [21]

Answer:

There are 1% probability that the last person gets to sit in their assigned seat

Step-by-step explanation:

The probability that the last person gets to sit in their assigned seat, is the same that the probability that not one sit in this seat.

If we use the Combinatorics theory, we know that are 100! possibilities to order the first 99 passenger in the 100 seats.

LIke we one the probability that not one sit in one of the seats, we need the fraction from the total number of possible combinations, of combination that exclude the assigned seat of the last passenger. In other words the amount of combination of 99 passengers in 99 seats: 99!

Now this number of combination of the 99 passenger in the 99 sets, divide for the total number of combination in the 100 setas, is the probability that not one sit in the assigned seat of the last passenger.

P = 99!/100! = 99!/ (100 * 99!) = 1/100

There are 1% probability that the last person gets to sit in their assigned seat

7 0
2 years ago
12. (2v^3 - v + 8) + (-v^3 + v - 3)<br><br> 14. (4h^3 + 3h + 1) - (-5h^3 + 6h - 2)
Lana71 [14]
12. On addition, you can just combine like terms.
2v^3+(-v^3)=v^3
-v+v cancels each other out
8+(-3)=5
So you have v^3+5

14 On subtraction, you have to remember to distribute the negative sign so after you do that you have:
4h^3+3h+1+5h^3-6h+2
Then you can combine like terms
4h^3+5h^3=9h^3
3h-6h=-3h
1+2=3
So you end up with:
9h^3-3h+3

Hope that helps and feel free to ask any questions.
7 0
3 years ago
I WILL GIVE BRAINLIEST
adelina 88 [10]

Answer: 2x - 1 (+x) =19

4 0
2 years ago
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