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klemol [59]
3 years ago
15

For the scientific notation, write the standard form: 5.61 x 102

Mathematics
1 answer:
Alekssandra [29.7K]3 years ago
4 0

<em>I believe the exponential form above is 5.61 x </em>10^{2}

Answer:

561

Step-by-step explanation:

A<em> "scientific notation"</em> is being used in order to express large numbers in its <u>simplest form</u>. This is commonly used in careers that use Mathematics very often like<em> engineers</em> or<em> scientists </em>because it makes their computation easier.

The<em> "standard form"</em> refers to the standard or traditional way of writing the numbers.

In order to get the standard form above, you have to check the <em>"power of the number 10." </em>The exponent above is 2. This means that you have to move the decimal point of 5.61 to the<em> right twice.</em> So, the answer is 561.

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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

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A rectangular parking lot has an area of 682 square yards. The lot is 22 yards wide. What is the length of the parking lot?
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The answer is 31 yards.
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3 years ago
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The equation f = v + at represents the final velocity of an object, f, with an initial velocity, v, and an acceleration rate, a,
omeli [17]
<span>f = v + at

Subtract v from both sides.

f - v = at

Divide both sides by a.

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Switch sides.

t = (f - v)/a

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3 years ago
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Please please help me!!
Alona [7]

Answer:

g(x) = f(x+1) + 1

Step-by-step explanation:

Since the graph is only moved you don't have to worry about the slope.

To move the graph to the left you add to the x and in this case, it only moved 1 over.

to move the graph up you add to the whole function and in this case, it only moved up 1.

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julio filled his car up with gas he paid $12.95 for 5 gallons of gas. what is the cost per gallon of gas
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Answer:

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Step-by-step explanation:

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