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djyliett [7]
3 years ago
10

How long would it take to make 4.5 million dollars with a yearly income of 1000 USD?

Mathematics
1 answer:
Angelina_Jolie [31]3 years ago
6 0

Answer:

4,500 days, years, or months

Your income is $1000, and you are trying to get $4,500,000.

You can do this is by dividing $4,500,000 with $1000.

<u>Division</u>

4500000/1000

Cross out all the numbers of 1000 into 4500000 until you get 1.

Since there are 3 zeros in 1000, you cross 3 out in 4,500,000

So, your answer will be 4,500.

It will take 4,500 days, years, or months (You didn't say the unit)

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X= 8
Y= -7

If you use elimination:

–3y–4x=–11
3y–5x=–61

Add the two equations and get

–9x=–72

Divide by –9 on both sides

X=8

Substitute 8 in for x

–3y–4(8)=–11

Simplify

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A school population was predicted to increase by 50 students a year for the next 10 years. If the current population is 700 stud
antoniya [11.8K]

The enrollment number should be 1200

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2 years ago
The outside of a picture frame measures 14 in and 20 in 196in^2 of the picture shows find the thickness of the frame
LuckyWell [14K]
The described picture frame can be visualized into two separate parts. The first area is equal to the area using the outermost dimensions for the length and width. 
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We are given that the area is only equal to 192 in². We subtract this value from the computed area.
         Difference = 280 in² - 192 in²
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This area is equal to the area of the hollow space inside the frame. That is equal to,
      height = 20 in - 2x
      length = 14 in - 2x

The area,
               88 = (20 - 2x)(14 - 2x)

Simplify the right hand side of the equation.
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Divide the equation by 4,
            22 = 70 - 17x + x²

Transposing,
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26-22x is the correct answer
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If D = Z + 6 and C = -2 + Z - Z^2, find an expression that equals 3D + C in standard form
kykrilka [37]

Answer:

4Z+16-Z²

Step-by-step explanation:

Given that,

D = Z+6

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We need to find the value of 3D+C.

So,

3D+C = 3(Z + 6)+(-2 + Z - Z²)

= 3Z+18-2+Z-Z²

= 4Z+16-Z²

So, the required expression is 4Z+16-Z².

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