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jenyasd209 [6]
3 years ago
8

4 different equations two have to be divide and two have to be multiply

Mathematics
1 answer:
wel3 years ago
6 0
768 ÷ 32 = 24
32 × 24 = 768

I hope this helps you!!
please make my answer brainliest to help me out!!!!
Thanks!!
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Kristen invests $ 5745 in a bank. The bank 6.5% interest compounded monthly. How long must she leave the money in the bank for i
podryga [215]

Answer:

1. 10.7 years

2. 17.0 years

3. 2nd option

Step-by-step explanation:

Use formula for compounded interest

A=P\cdot \left(1+\dfrac{r}{n}\right)^{nt},

where

A is final value, P is initial value, r is interest rate (as decimal), n is number of periods and t is number of years.

In your case,

1. P=$5745, A=2P=$11490, n=12 (compounded monthly), r=0.065 (6.5%) and t is unknown. Then

11490=5745\cdot \left(1+\dfrac{0.065}{12}\right)^{12t},\\ \\2=(1.0054)^{12t},\\ \\12t=\log_{1.0054}2,\\ \\t=\dfrac{1}{12}\log_ {1.0054}2\approx 10.7\ years.

2. P=$5745, A=3P=$17235, n=12 (compounded monthly), r=0.065 (6.5%) and t is unknown. Then


17235=5745\cdot \left(1+\dfrac{0.065}{12}\right)^{12t},

3=(1.0054)^{12t},

12t=\log_{1.0054}3,

t=\dfrac{1}{12}\log_{1.0054}3\approx 17.0\ years.

3. <u>1 choice:</u> P=$5745, n=12 (compounded monthly), r=0.065 (6.5%), t=5 years and A is unknown. Then

A=5745\cdot \left(1+\dfrac{0.065}{12}\right)^{12\cdot 5},\\ \\A=5745\cdot (1.0054)^{60}\approx \$7936.39.

<u>2 choice:</u> P=$5745, n=4 (compounded monthly), r=0.0675 (6.75%), t=5 years and A is unknown. Then

A=5745\cdot \left(1+\dfrac{0.0675}{4}\right)^{4\cdot 5},\\ \\A=5745\cdot (1.0169)^{20}\approx \$8032.58.

The best will be 2nd option, because $8032.58>$7936.39

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12

by-step explanation:

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An adult short-eared owl weighed 1/2 pound. A baby short-eared owl weighed 2/9 pound. What was the total weight of the two owls?
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Total weight of the two owls are 2/3
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When two lines intersect to form a 90 degree angle, these lines are called ________ lines?
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When two lines intersect to form a 90 degree angle, these lines are called perpendicular lines.
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