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Natasha_Volkova [10]
3 years ago
12

Rearrange the equation so m is the independent variable. -2m-5n=7m-3n n = __

Mathematics
2 answers:
WINSTONCH [101]3 years ago
8 0

Answer:n=-9/2m

Step-by-step explanation:

Step 1: Add 3n to both sides.

-2m-5n+3n=7m-3n+3n

-2m-2n=7m

Step 2: Add 2m to both sides.

-2m-2n+2m=7m+2m

-2n=9m

Step 3: Divide both sides by -2.

-2m/-2=9m/-2

n=-9/2

Hope I helped

MrRa [10]3 years ago
4 0

Answer:

n= -9m/2

Step-by-step explanation:

-2m-7m-5n=7m-7m-3n

-9m-5n+5n= -3n+5n

-9m=2n

n= -9m/2

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Two containers, A and B begin with equal volumes of liquid.
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Answer:

volume left in A is 80ml

Step-by-step explanation:

let original volume in both be x ml

after pouring from A to B

vol in A = x - 120

vol in B = x + 120

now it is given that

vol in B = 4 times vol in A

implies

x + 120 = 4 \times (x - 120) \\ x + 120 = 4x - 480 \\ 3x = 600 \\ x = 200

hence original volume in A is 200ml. after pouring in B it is 200-120 = 80 ml

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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
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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
3 years ago
How many more people voted for the most popular color than for the least popular color
PilotLPTM [1.2K]

Answer:

24

Step-by-step explanation:

red is most popular with 33 votes. silver is least popular with 9 votes.

33 - 9

24

7 0
3 years ago
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