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Svetllana [295]
2 years ago
10

Factor the greatest common factor (GCF) from each polynomial. 16x^4+24x^3-8x^2+12x​

Mathematics
1 answer:
Mila [183]2 years ago
8 0

Answer:

4x(4x^3+6x^2-2x+3)

Step-by-step explanation:

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Need help with solving please
attashe74 [19]

Answer:

x=8

Step-by-step explanation:

4 ^(1/2x) = 256

Rewriting 256 as a power of 4

4 ^(1/2x) = 4^4

Since the bases are the same, the exponents are the same

1/2x = 4

Multiply each side by 2

1/2x *2 = 4*2

x=8

5 0
3 years ago
Help please.........
WINSTONCH [101]

Lesser slope

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6 0
3 years ago
Ned started the month with $137.06 in his bank account. Over the course of the month he got 2 paychecks. One for $286.77 and the
jonny [76]

Answer:

Step-by-step explanation:

-228.91

7 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
Omar went to a Chicago Cubs game and purchased 1 pound of peanuts for $1.29. How much would 7 pounds of peanuts cost?
viktelen [127]

Answer:

$9.03

Step-by-step explanation:

if one pound of peanuts cost $1.29

and you want to know how much 7 pounds costs, you multiply $1.29 by 7.

$1.29×7=$9.03

3 0
3 years ago
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