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3 years ago

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The expression 1000(1.1)^t represents the value of a $1000 investment that earns 10% interest per year, compunded annually for t

years. What is the value of a $1000 investment at the end of each period? 1) 2 years 2) 4 years

1 answer:

Dmitriy789 [7]3 years ago

6 0

The answer is : $2 ~Year: A=1000(1.1)^2 A=$1210 For 4 Year: A=1000(1.1)^4 A=$1464.10$

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2 years ago

What is the solution to the inequality 10x + 18 < -2

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<h3>Answer: x < -2, choice B</h3>

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Work Shown:

10x + 18 < -2

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3 years ago

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2 years ago

Read 2 more answers

the product of two consecutive integers n and n+1 is 42. what is the positive integer that satisfies the situation?

MrMuchimi

$n(n+1)=42 \\
n^2+n=42 \\
n^2+n-42=0 \\ \\
a=1 \\ b=1 \\ c=-42 \\ b^2-4ac=1^2-4 \times 1 \times (-42)=1+168=169 \\ \\
n=\frac{-b \pm \sqrt{b^2-4ac}}{2a}=\frac{-1 \pm \sqrt{169}}{2 \times 1}=\frac{-1 \pm 13}{2} \\
n=\frac{-1 -13}{2} \ \lor \ n=\frac{-1+13}{2} \\
n=\frac{-14}{2} \ \lor \ n=\frac{12}{2} \\
n=-7 \ \lor \ n=6$

The positive integer is 6.
The consecutive integers are 6 and 7.

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3 years ago

The selling price of a certain video is $7.00 more than the price the store paid. If the selling price is $24.00, find the price

Ann [662]

The store paid $17.00

Solution: $24.00 (selling price) - $7.00 (difference of what the store paid) = $17.00

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2 years ago