So,
We start with 0 + 40.
1. 0 + 40
2. 1 + 39
3. 2 + 38
4. 3 + 37
...
...
...
19. 18 + 22
20. 19 + 21
21. 20 + 20
Tada! There are 21 pairs of whole numbers that have a sum of 40.
The distance between (3, 1) and (6, 5) is 5.
D=√(x₂-x₁)²+(y₂-y₁)²
D=√(6-3)²+(5-1)²
D=√3²+4²
D=√9+16
D=√25
D=5
Answer:
11a-8
Step-by-step explanation:
Please Note that i assume 12a and 2 are positive.
(12a+2−8a+4)-(12−7a+2)
=12a+2-8a+4-12+7a-2
=12a-8a+7a+2+4-12-2
=11a-8
Answer:
the coefficients are 4 and 5 and 1
Step-by-step explanation:
the 4 and 5 have a variable in front of them. There's a ghost 1 in front of the x, because it's value is 1 times whatever the value of x is
Answer:

Step-by-step explanation:

The fractions above are known as mixed numbers (that is, fractions that contain the combination of a whole number and proper fraction). The mixed numbers can equally be renamed as improper fractions. However, renaming can only occur the the mixed numbers are being transformed as below:

The transformation occur through a simple approach thus: for the first mixed number; 2 1/3 = 2 multiplied by 3 plus 1 divide by 3 equals 7/3. Similarly, mixed number 1 3/4 is transformed by this: 1 multiplied by 4 plus 3 divide by 4 equals 7/4.
Simplifying further to arrive at final result implies,

Take the LCM of the denominators (4 × 3 =12) and simplify thus:

