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Rus_ich [418]
2 years ago
8

Calculate the distance

Mathematics
1 answer:
andrey2020 [161]2 years ago
5 0

Answer:

Step-by-step explanation:

30/5=6

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Pls help me guys!<br> Thanks :D
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The top right is the answer. The base should be square with lengths of 8 and the height is 10.
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Ernest has 3/4 of a yard of ribbon. She used 1/2 of it to wrap a gift. How much ribbon did she use
Ivenika [448]


\frac{3}{4} (yard) \times \frac{1}{2} = \frac{3}{8} (yard) \ \\
5 0
2 years ago
X + 3 = x + 6<br> is it one solution no solution or infinitely solutions
mars1129 [50]

Answer:

No solutions I think.

Step-by-step explanation:

3 0
3 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
Jessica buys a lamp that costs $35 and a table that costs $75.00. If the tax rate is 10%, what will be the total cost of the lam
shepuryov [24]
Hello there!

75 + 35 = 110

110 x 10% = 0.1

110 x 0.1 = 11 ( the 11 is your tax )

75 + 35 = 110

110 + 11 = 121

$121 

Hope this helped!
3 0
3 years ago
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