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Murrr4er [49]
3 years ago
14

Solve for x: 3(x + 1) = −2(x − 1) + 6. (1 point) 1 4 5 25

Mathematics
1 answer:
Nikolay [14]3 years ago
3 0

Answer: 1

Step-by-step explanation:

You are welcome :)

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Triple the difference of 29 and 19
natta225 [31]

Answer:

30

Step-by-step explanation:

29 - 19 = 10 (difference)

triple 10 now:

10 x 3 = 30

7 0
1 year ago
jennifer has a taco stand. she has found that her daily costs can be modeled by c(x)=x^2-40x+610, where c(x) is the cost, in dol
Minchanka [31]
C(x) = 300.

Our question becomes simple: 300 = x^2-40x+610

x^2-40x+610-300 = 0
x^2-40x+310 = 0

Now find the value for x with either the quadratic formula or by factorization.

Let me know how it goes. Giving the full answer won’t help you improve
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
Find the volume r:7 h:7
sweet-ann [11.9K]

Answer:

V = 1077.566 units²

Step-by-step explanation:

Assuming this is a cylinder, the formula is (area of base) × (height)

or (πr²)(h), simply plug-in the given values:

= (π(7)²)(7)

= (153.938)(7)

V = 1077.566 units²

3 0
3 years ago
Is 9.37 greater or less than 5.999?
elixir [45]

Answer:

Greater

Step-by-step explanation:

The answer is greater because 9 is greater than 5. When working with decimals, always look at the whole number first!

8 0
3 years ago
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