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lawyer [7]
2 years ago
10

Sam can solve 30 multiplication problems in 2 mins. How many can he solve in 20 mins.

Mathematics
2 answers:
crimeas [40]2 years ago
8 0
To find the answer using unit rates, you do 30 divided by 2 which is 15/1. Then do 15*20 which is 300. So he can do 300 problems in 20 minutes.
Citrus2011 [14]2 years ago
6 0
To get to 20 minutes from 2 minutes you multiply by 10, do the same to the 30 problems to get 300. The answer is 300 problems in 20 minutes.
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A polynomial function g(x) with integer coefficients has a leading coefficient of 1 and a constant term of
mart [117]

Answer:

+1, -1, +11, -11

probably written with a +- symbol:

+-1, +-11

Maybe (silly) written like fractions:

+- 1/1, +- 11/1

Step-by-step explanation:

First list the factors of the leading coefficient. Here its 1. So we're going to use positive and negatives of the factors of 1, which is just +/- 1 . These numbers are going to go on the bottom of a fraction.

Next look for the factors of the constant, here it's 11

So that gives us

+/- 1, +/- 11 . These will go on the top of a fraction. (A fraction is a rational expression, that's why the name)

Then make all the combinations of

factors of constant

OVER

factors of leadingcoeff

So, we find

+/- 1, +/- 11

6 0
2 years ago
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Helen [10]
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8 0
3 years ago
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Maurinko [17]
Answer should be 22= 11x2 11= 11x1, something like that :)
5 0
2 years ago
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Sedaia [141]

Answer:

if X = -2 then

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8 0
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Read 2 more answers
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Salsk061 [2.6K]
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