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Marizza181 [45]
3 years ago
13

Please help ME!! I need help

Mathematics
1 answer:
viva [34]3 years ago
4 0

Answer:

You want to make it so you have an even amount of both, correct? This is where you need to know your multiples... here is what I got...

Step-by-step explanation:

By 4 packs of hotdogs and 5 packs of hotdog buns, this way you end up with 40 of each.

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Step-by-step explanation:

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If a trailer carries 625 gags of cocoa. How many bags will 265 trailers carry?
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3 0
3 years ago
25 (a square) - 4 (b square) + 28bc - 49 (c square)
LUCKY_DIMON [66]

Answer:

(5a - [2b - 7c]) and (5a + [2b + 7c])

Step-by-step explanation:

Factor 25a^2 - 4b^2 + 28bc - 49c^2.

Note that - 4b^2 + 28bc - 49c^2 involves the variables b and c, whereas 25a^2 has only one variable.  Thus, try to rewrite - 4b^2 + 28bc - 49c^2 as the square of a binomial:

- 4b^2 + 28bc - 49c^2 = -(4b^2 - 28bc + 49c^2), or

                                        -(2b - 7c)^2.

Thus, the original  25a^2 - 4b^2 + 28bc - 49c^2  looks like:

                               [5a]^2 - [2b - 7c]^2

Recall that a^2 - b^2 is a special product, the product of (a + b) and (a - b).  Applying this pattern to the problem at hand, we conclude:

Thus,   [5a]^2 - [2b - 7c]^2 has the factors (5a - [2b - 7c]) and (5a + [2b + 7c])

8 0
3 years ago
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