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Wewaii [24]
3 years ago
15

15 * Find the distance between the two points of G(-5, 4) and H(2,6). Round to the nearest tenth. (6 Points) 5.6 7.3 6.1 5.2​

Mathematics
1 answer:
madreJ [45]3 years ago
6 0

Answer:

7.3

Step-by-step explanation:

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Multiply the binomial and the trinomial. <br> (x-2)(x^2+2x+4)
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Answer:

Step-by-step explanation:

To solve this problem, we need to multiple x and -2 by x^{2} + 2x + 4 and add the results together:

x(x^{2} + 2x + 4)

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-2(x^{2} + 2x + 4)

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Adding the two together give the following:

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7 0
3 years ago
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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Find the next number in the sequence 32, 48, 56, 60, 62
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Answer: divide 2 from what you got from the last one

Step-by-step explanation:

32+16=48 so then divide 16 by 2 or subtract 8

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3 years ago
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