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Andreyy89
3 years ago
9

7/9÷1/9=2/3÷1/91/18÷1/91÷1/9

Mathematics
2 answers:
Paha777 [63]3 years ago
4 0
To divide fractions, take the second fraction and flip it upside down (step one)

Then multiply your two new fractions together (step two)

I showed you the first one, you can do the other ones! Really you can!!
Yuki888 [10]3 years ago
4 0

take the second fraction and flip it upside down like the dude said

Then multiply your two new fractions together

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ILL GIVE BRAINLIEST<br><br> evaluate the expression. 2•7(3^2)/3
scoray [572]

Answer:

\frac{7 \times  {3}^{2} }{3}  \\  \frac{7 \times 3 \times 3}{3}  \\  = 7 \times 3 = 21 \\ thank \: you

7 0
3 years ago
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7 0
2 years ago
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Two different cars each depreciate to 60% of their respective original values. The first car depreciates at an annual rate of 10
zvonat [6]

The approximate difference in the ages of the two cars, which  depreciate to 60% of their respective original values, is 1.7 years.

<h3>What is depreciation?</h3>

Depreciation is to decrease in the value of a product in a period of time. This can be given as,

FV=P\left(1-\dfrac{r}{100}\right)^n

Here, (<em>P</em>) is the price of the product, (<em>r</em>) is the rate of annual depreciation and (<em>n</em>) is the number of years.

Two different cars each depreciate to 60% of their respective original values. The first car depreciates at an annual rate of 10%.

Suppose the original price of the first car is x dollars. Thus, the depreciation price of the car is 0.6x. Let the number of year is n_1. Thus, by the above formula for the first car,

0.6x=x\left(1-\dfrac{10}{100}\right)^{n_1}\\0.6=(1-0.1)^{n_1}\\0.6=(0.9)^{n_1}

Take log both the sides as,

\log 0.6=\log (0.9)^{n_1}\\\log 0.6={n_1}\log (0.9)\\n_1=\dfrac{\log 0.6}{\log 0.9}\\n_1\approx4.85

Now, the second car depreciates at an annual rate of 15%. Suppose the original price of the second car is y dollars.

Thus, the depreciation price of the car is 0.6y. Let the number of year is n_2. Thus, by the above formula for the second car,

0.6y=y\left(1-\dfrac{15}{100}\right)^{n_2}\\0.6=(1-0.15)^{n_2}\\0.6=(0.85)^{n_2}

Take log both the sides as,

\log 0.6=\log (0.85)^{n_2}\\\log 0.6={n_2}\log (0.85)\\n_2=\dfrac{\log 0.6}{\log 0.85}\\n_2\approx3.14

The difference in the ages of the two cars is,

d=4.85-3.14\\d=1.71\rm years

Thus, the approximate difference in the ages of the two cars, which  depreciate to 60% of their respective original values, is 1.7 years.

Learn more about the depreciation here;

brainly.com/question/25297296

4 0
2 years ago
3/5x + 3 please help me out
Monica [59]
I’m sorry but there isn’t enough information here to answer anything. What does the equation equal?
5 0
3 years ago
Read 2 more answers
What is the value of X in the equation 2x+3y=36, when y=6?
zvonat [6]

Answer:

x = 9

Step-by-step explanation:

Substitute y=6 into 2x+3y=36

2x+3(6) =36

2x + 18 = 36

2x = 18

x = 9

7 0
3 years ago
Read 2 more answers
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