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kaheart [24]
3 years ago
10

Find a number such that 7 times the number is 56 greater than the opposite of the number.

Mathematics
2 answers:
lukranit [14]3 years ago
7 0

Answer:

4

Step-by-step explanation:

The opposite of a number is its negative. For example, the opposite of 3 would be -3.

Equipped with that information, we just need to find a number that is halve of 56, since if it stays the same except that its opposite, it will be 56 away.

Draw a number line. This can be a great help in problems like these.

If we divide 56 by 2, we get 28. Now the opposite of 28 is -28.

What happens if we add 56 to -28? We get positive 28.

Now that we know the number, we need to find out what number to multiply 7 by to get that.

The simple way to do this is to divide 28 by 7.

28/7=4.

7*4=28.

The answer is 4.

Hope this helps!

Elenna [48]3 years ago
6 0

Answer:

hey your answer can be 10 because the oppisite of 56 is 65 and 7x 10 is 70

Step-by-step explanation:

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Step-by-step explanation:

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Read 2 more answers
Need help Asap thank you in advance
julia-pushkina [17]

Answer:

12x^4-20x^2+32x=4x(3x^3-5x+8)

Step-by-step explanation:

<u>Common Factors</u>

An algebraic expression that is formed by sums or subtractions of terms can be factored provided there are numeric or variable common factors in all the terms.

The following expression

Z=12x^4-20x^2+32x

Can be factored in the constants and in the variable x.

1. To find the common factor of the variable, we must locate if the variable is present in all terms. If so, we take the common factor as the variable with an exponent which is the lowest of all the exponents found throughout the different terms. In this case, the lowest exponent is x (exponent 1).

2. To find the common factors of the constants, we take all the coefficients:

12 - 20 - 32

and find the greatest common divisor of them, i.e. the greatest number all the given numbers can be divided by. This number is 4, since 12/4=3, 20/4=5 and 32/4=8

3. The factored expression is

Z=12x^4-20x^2+32x=4x(3x^3-5x+8)

\boxed{12x^4-20x^2+32x=4x(3x^3-5x+8)}

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