Y=13
(when you have fractions like the one in this problem, you get rid of them by multiplying both sides by the denominator.)
The expression that is equivalent to 14xy – 28x – 36y + 48 is 2[7x(y-2)-6(3y-4)]
<h3>Factorizing expressions</h3>
Factorization is a way of separating the equations into two separate factors.
Given the expression below;
14xy – 28x – 36y + 48
Group
(14xy – 28x) – (36y + 48)
14x(y - 2) - 12(3y-4)
Factor out the value of 2 from both terms
2[7x(y-2)-6(3y-4)]
Hence the expression that is equivalent to 14xy – 28x – 36y + 48 is 2[7x(y-2)-6(3y-4)]
Learn more on factorization here: brainly.com/question/25829061
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Answer:
The factors of 2q²-5pq-2q+5p are (2q-5p) (q-1)....
Step-by-step explanation:
The given expression is:
2q²-5pq-2q+5p
Make a pair of first two terms and last two terms:
(2q²-5pq) - (2q-5p)
Now factor out the common factor from each group.
Note that there is no common factor in second group. So we will take 1 as a common factor.
q(2q-5p) -1(2q-5p)
Now factor the polynomial by factoring out the G.C.F, 2q-5p
(2q-5p) (q-1)
Thus the factors of 2q²-5pq-2q+5p are (2q-5p) (q-1)....
We can reject the last one: subtracting a non-zero value will result in a smaller value.
So now we have:
<span>2(A + B)
(A + B)2
A2 + B2
If all of them are mulptiplications, then they are all equivalent!
I mean by this, if what you meant is this:
</span>
<span>2*(A + B)
(A + B)*2
A*2 + B*2
If there is no sign, then the multiplication sign is implicit,
and all of these expressions say exactly the same: two of A and two of B.
</span>
