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pickupchik [31]
2 years ago
12

A structure covering a sports practice field has the shape of half of a cylinder, as shown. What is the total surface area of th

e structure to the nearest square yard, including the ends and the flat area of the field itself? Use 3.14 for .
Mathematics
1 answer:
hoa [83]2 years ago
3 0

wait i dont know why it wont let me comment or see comments but wheres the picture?

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When x is divided by 3 the quotient is more than 7
Yuri [45]
X/3 > 7 <== ur equation
multiply both sides by 3
x > 7 * 3
x > 21 <== ur solution
5 0
3 years ago
What is the Greatest common factor for -8X²+ 24​
monitta

Answer:

8

Step-by-step explanation:

- 8 {x}^{2}  + 24 \\  = 8( -  {x}^{2}  + 3) \\

5 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
Which expression are equivalent to -3-(7.5+4)
Hunter-Best [27]
-14.5

Basically, -(7.5+4) is -11.5 and then combine -3 with -11.5 which yields -14.5
3 0
3 years ago
157772+4426891-56 ajutor
arlik [135]

Answer:

4584607 is your correct answer

3 0
2 years ago
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