Simplify: negative two and one-third minus negative ten and one-sixth a negative twelve and one-half b twelve and one-half c seven and five-sixths d negative seven and five-sixths determine the answer to (−3) + (−5) and explain the steps using a number line. give example and how you did it use absolute value to express the distance between −12 and −15 on the number line. a |−12 − (−15)| = −27 b |−12 − (−15)| = −3 c |−12 − (−15)| = 3 d |−12 − (−15)| = 27 im failing and its stressing

Option C→Seven and five-sixths

2.

(-3) +(-5)

Mark -3 and -5 on the number line.Move 5 units left of -3 to reach at -8.That is required sum.

3.

Distance between −12 and −15 on the number line is given by

       |-12 -(-15}| or |-15 -(-12)|

=|-12+15| or |-15+12|

=|3| or |-3|

=3 units

Option C

|−12 − (−15)| = 3

d, x=8

step-by-step explanation:

simplifying

4x + -5 = 2x + 11

reorder the terms:

-5 + 4x = 2x + 11

reorder the terms:

-5 + 4x = 11 + 2x

solving

-5 + 4x = 11 + 2x

solving for variable 'x'.

move all terms containing x to the left, all other terms to the right.

add '-2x' to each side of the equation.

-5 + 4x + -2x = 11 + 2x + -2x

combine like terms: 4x + -2x = 2x

-5 + 2x = 11 + 2x + -2x

combine like terms: 2x + -2x = 0

-5 + 2x = 11 + 0

-5 + 2x = 11

add '5' to each side of the equation.

-5 + 5 + 2x = 11 + 5

combine like terms: -5 + 5 = 0

0 + 2x = 11 + 5

2x = 11 + 5

combine like terms: 11 + 5 = 16

2x = 16

divide each side by '2'.

step-by-step explanation:

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