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It is not necessary to have a common denominator when you multiply two fractions
<h3>How to determine the true statement?</h3>
The statement is about the relationship between the denominator of fractions when they are multiplied
Take for instance, we have the following fractions
a/b and x/y
In the above fractions, we have
Numerator = a and x
Denominator = b and y
Where b and y do not have the same value
When they are multiplied, we have
Product = ax/by
The denominators do not have the same value
And we were able to calculate the product
This means that the fraction denominator do not have to be common
Read more about fractions at
brainly.com/question/1622425
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You tutor 9 hours and you friend tutors for 3, i believe
Answer: 4n^2
Step-by-step explanation: -3n^2 - (-7n^2) = -3n^2 + 7n^2= 4n^2. When adding and subtracting the exponents stay the same only the coefficients are subtracted or added
Step-by-step explanation:
The equation will form a circle because it in the form of

First, let set the equation equal.

Here the center will be (7,-5), and the radius of 5. The boundary line will be dashed
Since this is in the inequalities, we must find the solution set.
Plug in 0,0 for x and y and see if it's true.



This is a true so we shade the region that includes 0,0
Since 0,0 has a greater distance from the center of the circle, 0,0 is outside of the circle, so our solution set will
be outside of circle.
Here a picture of graph,