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Veseljchak [2.6K]
3 years ago
7

If you answer this question correctly i'll give you brainliest.

Mathematics
2 answers:
Svet_ta [14]3 years ago
8 0

Answer:

first 11 second is 12 and 13 third is 14

Step-by-step explanation:

Softa [21]3 years ago
5 0

Answer:

5.7.11.13

Step-by-step explanation:

My guess

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Convert to an improper fraction. Type in your answer with the negative in the numerator.
MariettaO [177]

Answer:

-35/4

Step-by-step explanation:

6 0
3 years ago
What is:<br> 13 to the 3rd power<br> 2/5 to the second power<br> And 0.9 to the second power
Anna007 [38]
1. 13*13*13 =2197
2. 2/5*2/5 = 4/25
3. 0.9*0.9 = 0.81
4 0
3 years ago
Read 2 more answers
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
1/8 multiplied by 4 2/3
Anna35 [415]
First you Change the mixed number to an improper fraction (14/3) then you simplify (1/4 X 7/3) and then you multiply 1/4 X 7/3 = 7/12 :)
5 0
3 years ago
Read 2 more answers
Tia walked 1/2 mile in 5 minutes. She walked at the same rate for the entire distance. How far did Tia walk in 1 minute?
Hatshy [7]
She walked 1/10 of a mile in one minute
work: 1/2 a mile is 0.5 and 0.5 divided by 5 is 0.1 so you get 0.1 mile for each minute
3 0
3 years ago
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