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evablogger [386]
3 years ago
9

Jill has 15 oranges and 12 pears. She wants to put all of the fruit into baskets with each basket having the SAME NUMBER of piec

es of fruit in it. Without mixing the fruit, what is the GREATEST number of pieces of fruit Jill can put in each basket?
Mathematics
1 answer:
aleksandr82 [10.1K]3 years ago
8 0
Y’all need help ASAP thanks for
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From the top of a cliff, which is 120 meters above the water, tha angle of depression of a boat on the water is 18°. How far is
harina [27]
To find how far the boat is from the bottom of the cliff is you’ll use tan;

Tan18degrees=120/x

x=120/tan18degrees

x=369m
4 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

4 0
2 years ago
Write 0.01 as a fraction<br><br> a. 1 1/10<br> b. 1/10<br> c. 1/100<br> d. 1
Lady_Fox [76]
C. 01/100.    .01 is in the hundredths place. So you put 2 0's on the denominator.
5 0
3 years ago
Read 2 more answers
6) Sally wrote the number 30, 048 in expanded form.
vichka [17]

Answer:

the answer is G) Change 30 to 3,000

6 0
3 years ago
A rectangular school banner has a length of 40 inches and a width of 24 inches. A sign is made that is similar to the school ban
e-lub [12.9K]

Answer:

The answer is 960

Step-by-step explanation:all you would have to do is multiply the lenth times width. 40x24=960  


4 0
3 years ago
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