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iren [92.7K]
2 years ago
10

Which expression is an equivalent expression?

Mathematics
2 answers:
Natalka [10]2 years ago
8 0
So, first you would distribute 2 and 1/4 which is 1/2z then do 2 times 8 which is 16 and have it as 1/2z-16 then take 1/4z and add it to 1/2 since your adding the two expressions. So whatever 1/4z plus 1/2z is then you subtract is to 16 minus 6, because 6 is in your other expression. It's kinda hard to explain but the answer is C haha Hope this helped! 
VikaD [51]2 years ago
8 0

Answer:

The equivalent of the given expression: 2(¼x - 8) + (¼x + 6) is ¾x - 10

Step-by-step explanation:

Given.

Expression: 2(¼x - 8) + (¼x + 6)

To get an equivalent expression, we have to simplify the above expression

2(¼x - 8) + (¼x + 6)

First, open the brackets.

2 * ¼x - 2 * 8. + ¼x + 6

Perform multiplication operations

½x - 16 + ¼x + 6

Collect like terms

½x + ¼x - 16 + 6

Take LCM of ½x and ¼x (LCM is 4); then solve

(2x + x)/4 - 16 + 6.

Solve bracket

3x/4 - 16 + 6

¾x - 10

Hence, the equivalent of the given expression: 2(¼x - 8) + (¼x + 6) is ¾x - 10

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Answer:

Step-by-step explanation:

To solve these equations involving variables and exponents we need to follow these steps.

1) We need to find out the factor that is common in the equation.

2) After taking common, solve the equation. We can add or subtract only those values that have same bases.

1) 8+6x^4

here we can see, both numbers are divisible by 2, so taking 2 common

=2(8/2 + 6x^4/2)\\= 2(4 + 3x^4)

It cannot be further simplified because both number donot have same bases.

3.4n^9 + 12 n

We can take 4n common

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5. -12a -3

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= -3(-12a/-3 -3/-3)

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here the smallest power of n is n^3 so, we can take n^3 common and both coefficients are divisible by 4 so taking 4n^3 common

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Here we cannot take k common, as k is not a multiple of 10. For taking common it should be divisible by each value in the equation. But each value s divisible by 5 so, taking 5 common

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Here we cannot take n common, as n is not a multiple of -60. For taking common it should be divisible by each value in the equation. But each value s divisible by 10 so, taking 10 common

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Here we cannot take n common, as n is not a multiple of 28. For taking common it should be divisible by each value in the equation. But each value s divisible by -4 so, taking -4 common

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