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kifflom [539]
3 years ago
8

Jennifer receives $100 and puts it into her savings account. She adds $0.25 to the account each day for a number of days, d, aft

er that. She writes the expression 100 + 0.25(d – 1) to find the total amount of money in her savings account after ddays. Which statement about her expression is true?
A.
It is the sum of the initial amount and the additional amount after d days.

B.
It is the product of the initial amount and the additional amount after d days.

C. It is the sum of the initial amount and the number of days, 

D.
It is the difference between the initial amount and the additional amount after d days
Mathematics
2 answers:
PilotLPTM [1.2K]3 years ago
7 0

Answer:

B.) It is the product of the initial amount and the additional amount after d days.


Step-by-step explanation:

You multiply in the problem to get your answer.


kkurt [141]3 years ago
6 0

Answer:

A. It is the sum of the initial amount and the additional amount after d days.

Step-by-step explanation:

We have been given an expression 100+0.25(d-1), which gives the total amount of money in Jennifer's savings account after d days.

Since Jennifer adds $0.25 each day after first day, so amount in Jennifer's account after d days would be 100 plus 0.25 times d minus 1 day.

Since Jennifer "adds" an amount each day, therefore, the expression is the sum of initial amount and the additional amount after d days.

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S = 3.500t

11 weeks, because in 11 weeks he will have saved up 38.500, + the 5 he has saved equals 43.500.

Step-by-step explanation:

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4 0
3 years ago
Find the distance, in feet, a particle travels in its first 2 seconds of travel, if it moves according to the velocity equation
just olya [345]

Answer:

The particle will travel 6 feet in first 2 seconds.

Step-by-step explanation:

We have been given that a particle moves according to the velocity equation v(t)= 6t^2-18t+12. We are asked to find the distance that the particle will travel in its first 2 seconds.

s(t)=\int |v(t)|dt

s(t)=\int\limits^2_0 |6t^2-18t+12|dt

Now, we will eliminate the absolute value sign as:

s(t)=\int\limits^1_0 6t^2-18t+12dt+\int\limits^2_1 -6t^2+18t-12dt

s(t)=[\frac{6t^3}{3}-\frac{18t^2}{2}+12t]^1_0 +[\frac{-6t^3}{3}+\frac{18t^2}{2}-12t]^2_1

s(t)=[2t^3-9t^2+12t]^1_0 +[-2t^3+9t^2-12t]^2_1

s(2)=2(1)^3-9(1)^2+12(1)-(2(0)^3-9(0)^2+12(0))-2(2)^3+9(2)^2-12(2)-(-2(1)^3+9(1)^2-12(1))

s(2)=2-9+12-(0)-16+36-24-(-2+9-12)

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s(2)=5-4+5

s(2)=6

Therefore, the particle will travel 6 feet in first 2 seconds.

   

 

7 0
3 years ago
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