This is a difference of squares because the number in front of x2 is a perfect square and the constant, 9, is also a perfect square. Just take the square root of both numbers and write it as follows ...
This is a difference of squares because the number in front of x2 is a perfect square and the constant, 9, is also a perfect square. Just take the square root of both numbers and write it as follows ...
Everywhere you see an "n", replace it with a -6, so -6/6 - 3(-6) + 10 -1 + 18 + 10 27
Please rephrase your question. It doesn't make a lot of sense. There are two theorems for triangle congruency that involve two congruent angles. These are Angle, angle, Side and Angle, Side, angle. Once again, please clarify your question and I'll try to help you more. Thanks!
First factor out the n and a 8 because n and 8 go into each part. This gives you 8n (3n2 - 5n + 9). Always look for a common factor you can get rid of. The quadratic inside the parenthesis does not factor so this is your final answer.
Okay, first, let's get the z/6 by itself so we need to get rid of the -11. Add -11 to both sides to get z/6 = 11 Now, to get rid of the denominator you need to undo division by 6. The opposite of division by 6 is multiplication by 6. Let's undo the division by 6 by multiplying both sides...
From your question there is two ways to interpret it. Is the +1 in the denominator with the 3 or is it separate from the fraction? If the +1 is separate from the fraction, turn it into a fraction with a 3 in the denominator so you can add it to the x/3. So, 1 becomes 3/3. That whole...
First look at the whole numbers and find a common factor. In this case, 2 goes into all of those numbers so we will pull a 2 out and divide all the numbers by 2. 2(6cd3 - 4c2d2 + 5c5d3) Now we look to see if everything has a c and if so, which c has the smallest exponent. The first c...
What we want to do here is first check to see if the problem is organized where terms that have common factors are together. This problem meets that criteria because the first 2 terms have x and the 3rd and 4th terms have y in them. Next, we write what both have in common first, so x You...