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lisov135 [29]
3 years ago
15

For 50 Points!!!!!!!!!

Mathematics
2 answers:
dolphi86 [110]3 years ago
5 0
Hello!

First you have to find what the temperature was at the 3 stages

It started at 87

Then went down 10

87 - 10 = 77

Then up 17

77 + 17 = 94

Then you take the difference from the current temperature from the lowest temperature

94 - 77 = 17

The answer is D

Hope this helps!
Brut [27]3 years ago
5 0
The answer is 17 i took the test
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Where is the midpoint of AB if A(6,4) and B(5,-2)?
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Answer:

Midpoint={5.5,1}

Step-by-step explanation:

M={x1+x2/2,y1+y2/2}

M={6+5/2,-2+4/2}

M={11/2,2/2}

M={5.5,1}

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If a line crosses the y-axis at (0, 1) and has a slope of 4/5 , what is the equation of the line?
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45 miles per hour into feet per second please show work I am hopless :(
Pavel [41]
There are 5280 feet in a mile
45 * 5280 = 237600
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5 0
3 years ago
Solve -11 2/3 * (-4 1/5)
eduard

Answer:pts

Solve -11 2/3 * (-4 1/5)=49

Step-by-step explanation:

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mark me brainest plz

6 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
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