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erma4kov [3.2K]
3 years ago
5

Kenzie got a gift card for $30.00 to spend in the app store. Each app cost d dollars. write an expression to represent the amoun

t of money Kenzie had left on her gift card after she bought 6 apps.
Mathematics
2 answers:
dimaraw [331]3 years ago
8 0
The correct equation would be y=30-6d, the y would be the amount of money left.
Vinvika [58]3 years ago
7 0
Y=30d-6 that’s ur answer
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Help please!! I'm trying to solve this question but I can't seem to find the length of one side properly, that said, I'm unable
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Assuming the rectangles are squares: Length and width are equal so

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5 0
3 years ago
A 20-year loan of 1000 is repaid with payments at the end of each year. Each of the first ten payments equals 150% of the amount
Alja [10]

Answer:

x = 97

Step-by-step explanation:

Given

t = 20 --- time (years)

A =1000 --- amount

r = 10\% --- rate of interest

Required

The last 10 payments (x)

First, calculate the end of year 1 payment

y_1(end) = 10\% * 1000 * 150\%

y_1(end) = 150

Amount at end of year 1

A_1=A - y_1(end) - r * A

A_1=1000 - (150 - 10\% * 1000)

A_1 =1000 - (150- 100)

A_1 =950

Rewrite as:

A_1 = 0.95 * 1000^1

Next, calculate the end of year 1 payment

y_2(end) = 10\% * 950 * 150\%

y_2(end) = 142.5

Amount at end of year 2

A_2=A_1 - (y_2(end) - r * A_1)

A_2=950 - (142.5 - 10\%*950)

A_2 = 902.5

Rewrite as:

A_2 = 0.95 * 1000^2

We have been able to create a pattern:

A_1 = 1000 * 0.95^1 = 950

A_2 = 1000 * 0.95^2 = 902.5

So, the payment till the end of the 10th year is:

A_{10} = 1000*0.95^{10}

A_{10} = 598.74

To calculate X (the last 10 payments), we make use of the following geometric series:

Amount = \sum\limits^{9}_{n=0} x * (1 + r)^n

Amount = \sum\limits^{9}_{n=0} x * (1 + 10\%)^n

Amount = \sum\limits^{9}_{n=0} x * (1 + 0.10)^n

Amount = \sum\limits^{9}_{n=0} x * (1.10)^n

The amount to be paid is:

Amount = A_{10}*(1 + r)^{10} --- i.e. amount at the end of the 10th year * rate of 10 years

Amount = 1000 * 0.95^{10} * (1+r)^{10}

So, we have:

Amount = \sum\limits^{9}_{n=0} x * (1.10)^n

\sum\limits^{9}_{n=0} x * (1.10)^n = 1000 * 0.95^{10} * (1+r)^{10}

\sum\limits^{9}_{n=0} x * (1.10)^n = 1000 * 0.95^{10} * (1+10\%)^{10}

\sum\limits^{9}_{n=0} x * (1.10)^n = 1000 * 0.95^{10} * (1+0.10)^{10}

\sum\limits^{9}_{n=0} x * (1.10)^n = 1000 * 0.95^{10} * (1.10)^{10}

The geometric sum can be rewritten using the following formula:

S_n = \sum\limits^{9}_{n=0} x * (1.10)^n

S_n =\frac{a(r^n - 1)}{r -1}

In this case:

a = x

r = 1.10

n =10

So, we have:

\frac{x(r^{10} - 1)}{r -1} = \sum\limits^{9}_{n=0} x * (1.10)^n

\frac{x((1.10)^{10} - 1)}{1.10 -1} = \sum\limits^{9}_{n=0} x * (1.10)^n

\frac{x((1.10)^{10} - 1)}{0.10} = \sum\limits^{9}_{n=0} x * (1.10)^n

x * \frac{1.10^{10} - 1}{0.10} = \sum\limits^{9}_{n=0} x * (1.10)^n

So, the equation becomes:

x * \frac{1.10^{10} - 1}{0.10} = 1000 * 0.95^{10} * (1.10)^{10}

Solve for x

x = \frac{1000 * 0.95^{10} * 1.10^{10} * 0.10}{1.10^{10} - 1}

x = 97.44

Approximate

x = 97

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3 years ago
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Answer:

10 - 10 as in 10 minus 10? 10 minus 10 is 0. Both of you are wrong.

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4 years ago
What is the range of this promblem
castortr0y [4]

Answer:

3

Step-by-step explanation:

Range is the largest number minus the smallest number

4-1 = 3

The range is 3

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3 years ago
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