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sergey [27]
2 years ago
9

In order to make the expression below equivalent to 1 2 x 6, which additional operation should be included in the expression? Us

e the drop-down menu to select the operation. Five-fourths x 6.
Mathematics
1 answer:
marysya [2.9K]2 years ago
7 0

The additional operation which needed to make the expression equivalent to given expression is -\dfrac{3}{4} x. Thus the option 3 is the correct option.

<h3>What is equivalent expression?</h3>

Equivalent expression are the expression whose result is equal to the original expression, but the way of representation is different.

Given information-

The equation given in the problem is,

\dfrac{5}{4} x+6

Let the equation is equation number 1.

This expression is to convert in the expression, which is,

\dfrac{1}{2}x+6

Let the above equation is equation number 2.

Let the additional operation which needed to make the expression equivalent to given expression is f(x).

As the addition of f(x) and equation 1 is equal to the equation 2. Therefore,

f(x)+\dfrac{5}{4} x+6=\dfrac{1}{2}x+6

Solve the equation as,

f(x)=\dfrac{1}{2}x+6-\dfrac{5}{4} x-6\\f(x)=\dfrac{1}{2}x-\dfrac{5}{4} x\\f(x)=\dfrac{2-5}{4} x\\f(x)=\dfrac{-3}{4} x

Hence, the additional operation which needed to make the expression equivalent to given expression is -\dfrac{3}{4} x. Thus the option 3 is the correct option.

Learn more about the equivalent expression here;

brainly.com/question/2972832

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In a marble collection, 1/8 of the marbles are blue. Of the blue marbles, 1/2 have sparkles. What fraction of the marbles in the
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kvasek [131]

When you are multiplying an exponent directly into a number/variable with an exponent, you multiply the exponents together.

For example:

(x^{2} )^{3} = x^6

(x^{3} )^5=x^{15}


When you are multiplying a variable with an exponent by another variable with an exponent, you add the exponents together.

For example:

(x^{2} )(x^{3})=x^{5}

(x^{1} )(x^{2})=x^{3}


(\frac{(x^{-3})(y^{2})}{(x^{4})(y^{6})} )^{3}=\frac{(x^{-9})(y^{6})}{(x^{12})(y^{18})}

You multiply 3 into each exponent in the numerator and the denominator

\frac{(x^{-9})(y^{6})}{(x^{12})(y^{18})}= \frac{y^{6}}{(x^{9})(x^{12})(y^{18})}

When you have a negative exponent, you move it to the other side of the fraction to make the exponent positive.

\frac{y^{6}}{(x^{21})(y^{18})} = \frac{1}{(x^{21})(y^{12})}


When you have something like this:

\frac{x^{2}}{x^5}

You subtract the exponents together, so:

\frac{x^2}{x^5} = x^{2-5} = x^{-3} = \frac{1}{x^3}


Your answer is the second option

3 0
3 years ago
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