Answer:
Part 1)
-----> 
Part 2)
------> 
Part 3)
-----> 
Part 4)
-----> 
Part 5)
------> 
Step-by-step explanation:
Part 1) we have

To calculate the subtraction convert the mixed numbers to an improper fractions



substitute

Convert to mixed number

Part 2) we have

To calculate the subtraction eliminate the parenthesis first

Part 3) we have

To calculate the subtraction convert the mixed numbers to an improper fractions



substitute

Eliminate the parenthesis

Convert to mixed number

Part 4) we have

To calculate the subtraction eliminate the parenthesis first

Part 5) we have

To calculate the subtraction convert the mixed numbers to an improper fractions



substitute

Eliminate the parenthesis

Answer:
= 105(x+2)/33-x
Step-by-step explanation:
Given the expression
[3(x+2)*10*7]÷70 - 2(x+2)
= 3(x+2)*70÷70 - 2(x+2)
= 210(x+2)/70-2x-4
= 210(x+2)/66-2x
= 210(x+2)/2(33-x)
= 105(x+2)/33-x
Answer:
See Explanation
Step-by-step explanation:
a) Additive inverse of −2
- the additive inverse of a number a is the number that, when added to 'a', yields zero. This number is also known as the opposite (number), sign change, and negation.
- So the Additive inverse of -2 is 2. ∴ -2+2=0
b) Additive identity of −5
- Additive identity is the value when added to a number, results in the original number. When we add 0 to any real number, we get the same real number.
- -5 + 0 = -5. Therefore, 0 is the additive identity of any real number.
c) additive inverse of 3
- Two numbers are additive inverses if they add to give a sum of zero. 3 and -3 are additive inverses since 3 + (-3) = 0. -3 is the additive inverse of 3.
d). multiplicative identity of 19
- an identity element (such as 1 in the group of rational numbers without 0) that in a given mathematical system leaves unchanged any element by which it is multiplied
- Multiplicative identity if 19 is 1 only, since 19 x 1 = 19.
e) multiplicative inverse of 7
- Dividing by a number is equivalent to multiplying by the reciprocal of the number. Thus, 7 ÷7=7 × 1⁄7 =1. Here, 1⁄7 is called the multiplicative inverse of 7.
d) | 11-5|×|1-5|
- | 11-5|×|1-5| ⇒ I6I×I-4I ⇒ 6×4 ⇒ 24