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Alex Ar [27]
4 years ago
10

Kim learned in her science class that every 2 minutes she spends in the shower, she uses 17 gallons of water. This rate is const

ant
Mathematics
1 answer:
exis [7]4 years ago
6 0
The ratio should be 2 /17 
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If Jay ran 1 mile in 10 minutes on Monday, 2 miles in 20 minutes on Tuesday,
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Answer:

4 miles

Step-by-step explanation:

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4 years ago
Which equation is equivalent to<br> 5 - 2x = 4?
expeople1 [14]
-2x = -1
x = 1/2
Any equation that returns x=1/2 is a solution to this problem. For example, x= (3•4)/6 can be simplified so that x=1/2
5 0
3 years ago
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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

4 0
2 years ago
Someone please help I really appreciate it
Alex_Xolod [135]
So you have to find the relationship between the pumpkin diameter (y) and the number of weeks passed (x) in a table, lets do it taking into account that the equation modeling such behaviour is:
y = 2x + 6, where x is the number of weeks and plus original 6 cm
x                                y                            
week                         diameter
0                                6
1                                8
3                                12
5                                16
10                               26
substitute the x and y values in the equation to see how they fit into it
5 0
3 years ago
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How do you expand brackets
Olenka [21]

Answer:

multiply each term in the bracket by the expression outside the bracket

:)

Step-by-step explanation:

7 0
3 years ago
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