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yKpoI14uk [10]
3 years ago
13

How many weeks are in 3 years​

Mathematics
2 answers:
kondor19780726 [428]3 years ago
6 0
156.53 weeks are in 3 years
ycow [4]3 years ago
5 0

Answer:

156.429

Step-by-step explanation:

52.1428571 in one year x 3

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Tell whether the equation has one, zero, or infinitely many solutions:<br>9(x - 4) +15=9x – 21​
Mama L [17]

Answer:

Infinite solutions

Step-by-step explanation:

Simplify both sides, subtract 9x, and add 21x to both sides and you will get 0 = 0 so there is infinite solutions.

8 0
3 years ago
8x plus 8y equals 8. 8x plus 2y equals 28. Find x and y
alukav5142 [94]

Answer:

x,y = 13/3, -10/3

Step-by-step explanation:

8x+8y=8

8x+2y=28  Subtract eq1 by eq 2

6y=-20

y=-20/6 y = -10/3

8x - 80/3=8

8x=34 2/3

x=4 1/3

x=13/3

Hope this helps plz hit the crown :D

6 0
3 years ago
Read 2 more answers
I need help! please!
mamaluj [8]

Answer:

-  \frac{1}{3}

Step-by-step explanation:

The only thing important in this question in the original equation is the slope, which is 3, the multiplyer for x. To find the opposite slope follow the steps below.

The first step is to make the original slope into a fraction.

\frac{3}{1}

Then flip the numerator and denominator.

\frac{1}{3}

Now multiply this value by -1.

\frac{1}{3}  \times ( - 1) =  -  \frac{1}{3}

There is the opposite slope of 3.

7 0
2 years ago
Find the quotient for 5,600÷8​
rodikova [14]

Answer:

700

Hope this helps!!

7 0
3 years ago
Read 2 more answers
Simplify each only using positive exponents:<br> 2x^-3 • 4x^2<br> 2x^4 • 4x^-3<br> 2x^3y^-3 • 2x
erica [24]
<h2>Answer:</h2>

\frac{2}{x}

\frac{x}{2}

\frac{4x^4}{y^3}

<h2>Step-by-step explanation:</h2>

a. 2x^-3 • 4x^2

To solve this using only positive exponents, follow these steps:

i. Rewrite the expression in a clearer form

2x⁻³ . 4x²

ii. The position of the term with negative exponent is changed from denominator to numerator or numerator to denominator depending on its initial position. If it is at the numerator, it is moved to the denominator. If otherwise it is at the denominator, it is moved to the numerator. When this is done, the negative exponent is changed to positive.

In our case, the first term has a negative exponent and it is at the numerator. We therefore move it to the denominator and change the negative exponent to  positive as follows;

\frac{1}{2x^3} . 4x^2

iii. We then solve the result as follows;

\frac{1}{2x^3} . 4x^2 = \frac{2}{x}

Therefore, 2x⁻³ . 4x² = \frac{2}{x}

b. 2x^4 • 4x^-3

i. Rewrite as follows;

2x⁴ . 4x⁻³

ii. The second term has a negative exponent, therefore swap its position and change the negative exponent to a positive one.

2x^4 . \frac{1}{4x^3}

iii. Now solve by cancelling out common terms in the numerator and denominator. So we have;

\frac{x}{2}

Therefore, 2x⁴ . 4x⁻³ = \frac{x}{2}

c. 2x^3y^-3 • 2x

i. Rewrite as follows;

2x³y⁻³ . 2x

ii. Change position of terms with negative exponents;

2x^3.\frac{1}{y^3} .2x

iii. Now solve;

\frac{4x^4}{y^3}

Therefore, 2x³y⁻³ . 2x = \frac{4x^4}{y^3}

8 0
3 years ago
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