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soldier1979 [14.2K]
3 years ago
11

Daniel borrows $2,500 at 6% simple interest for 5 years. How much is daniel going to pay in interest. May I please get help I ha

ve already tried and it is to hard for me.
Mathematics
1 answer:
Sergio039 [100]3 years ago
5 0

Answer:

2500 x 6/100 = 150

150 x 5 = 750

2500 + 750 = 3 250

Step-by-step explanation:

You first have to determine the interest for one year. We know that a percentage means a 100 so we are going to use 6/100. You then have to multiple that by the original amount, you will then get your answer. After obtaining your answer your will multiple it by the number of years which is 5, once you get your answer add it to the original amount. Then you are all set.

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4 years ago
Janet and Nadia each play basketball. Nadia has won twice the number of games Janet has. Is it possible for Janet to have won 10
ladessa [460]

Answer: It is not possible for Janet to have won 10 games.

Step-by-step explanation:

Let be "x" the number of games Janet won.

We know that Nadia has won twice the number of games Janet has and the sum of the games Nadia and Janet have won together is 24.

Then, we express this situation with this equation:

x+2x=24

So, let's check if it is possible for Janet to have won 10 games. Substitute x=10 into the expression:

10+2(10)=24\\\\10+20=24\\\\30=24\ (This\ is\ not\ true)

Therefore, it is not possible for Janet to have won 10 games.

3 0
4 years ago
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Factorize x^3-2x^2-13x-10
Alecsey [184]
X^3-2x^2-13x-10=(x+1)(x+2)(x-5)
4 0
3 years ago
Martin is making squares by arranging 26 sticks. how many squares can he make? Write and solve the number fact you used to find
Alexxx [7]
He could make 6 squares, because each square has 4 sides. And 24 divided by 4 is 6. The number fact that i used is 24 divided by 4
6 0
3 years ago
I need to verify identity functions for a one to one function.
Eddi Din [679]

We have the function

f(x)=(x+6)^3

1. For f^-1:

Let y = f(x) = (x+6)^3

Switch x and y to get:

x=(y+6)^3

And solve for y

\begin{gathered} x^{\frac{1}{3}}=y+6 \\ x^{\frac{1}{3}}-6=y+6-6 \\ x^{\frac{1}{3}}-6=y \end{gathered}

And we have y = f^-1(x)

Answer blank 1:

f^{-1}(x)=x^{\frac{1}{3}}-6

2. For f o f^-1 (x):

(f\circ f^{-1})(x)=f(f^{-1}(x))

And solve

\begin{gathered} =f(x^{\frac{1}{3}}-6) \\ =(x^{\frac{1}{3}}-6+6)^3 \\ =(x^{\frac{1}{3}})^3 \\ =x \end{gathered}

answer blank 2

x^{\frac{1}{3}}-6

answer blank 3

x^{\frac{1}{3}}-6

answer blank 4

x^{\frac{1}{3}}

3. For f^-1 o f:

(f^{-1}\circ f)(x)=f^{-1}(f(x))

Solve

\begin{gathered} =f^{-1}((x+6)^3) \\ =\sqrt[3]{(x+6)^3}-6 \\ =x+6-6 \\ =x \end{gathered}

answer blank 5

(x+6)^3

answer blank 6

(x+6)^3

answer blank 7

x+6

4 0
1 year ago
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