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dem82 [27]
3 years ago
7

I’ve posted this a lot already and i keep getting it wrong can you help pls?

Mathematics
1 answer:
Schach [20]3 years ago
5 0

Answer: See below; x=1/6

Step-by-step explanation:

This problem may seem intimidating, but all you need to know are your exponent rules. Let's first work inside the parenthesis. Since the bases are the same, we can directly subtract the exponents.

\frac{3}{4} -\frac{3}{8}                             [common denominator]

\frac{6}{8} -\frac{3}{8}                             [subtract]

\frac{3}{8}

Now, our new answer is (3^\frac{3}{8} )^\frac{4}{9}.

Unfortunately, the answer is asking for an exponent with a single base. One of the properties is when the exponents are separated by parenthesis, you multiply them together.

\frac{3}{8} *\frac{4}{9}                              [cross cancel to simplify]

\frac{1}{2} *\frac{1}{3}                              [multiply]        

\frac{1}{6}

Our final answer is 3^\frac{1}{6}.

The value is x is \frac{1}{6} since 3^\frac{1}{6} is equal to 3^x, we know that x=\frac{1}{6}.

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Step-by-step explanation:

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What is the solution of this system of linear equations? 3y=3/2x+6 1/2y-1/4x=3.
tensa zangetsu [6.8K]

Answer:

No solution

Step-by-step explanation:

Please, separate equations with a comma, or the word "and," or a semicolon.

Next, determine the LCD and multiply both equations by it, so as to elimiinate the fractional coefficients.

3y=3/2x+6 1/2y-1/4x=3

should be written as the system

   3y=(3/2)x+6         Use parentheses around the fraction for clarity

(1/2)y-(1/4)x=3            Same:  use parentheses

The LCDs here are 2 (for the first equation) and 4 (for the second equation).  Multiply the first equation by 2 to eliminate the fractions:

   3y=(3/2)x+6         →   6y = 3x + 12, and

4[ (1/2)y-(1/4)x=3 ]     →   2y  -  x   = 12

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6y = 3(2y - 12) + 12, or

6y = 6y - 36 + 12

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this system has NO solution.

 

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