Answers/Solutions:
26.) 17ab² - 3ab
First, subtract (-5a²b+7ab²) and (a²b+9ab²).
-5a²b+7ab² - a²b + 9ab² =
(-5a²b - a²b) + (7ab² + 9ab²) =
-4a²b + 16ab²→ difference
Then, we must add the difference from 4a²b - 3ab + ab².
(4a²b - 3ab + ab²) + (-4a²b + 16ab²) =
(4a²b - 4a²b) -3ab + (ab² + 16ab²) =
-3ab + 17ab²
Final Answer: 17ab² - 3ab
27.) 3x² + 7y² + 26
First, add (7x² + 5y² + 8) and (-y² + 11).
7x² + 5y² + 8 -y² + 11 =
7x² + (5y² - y²) + (8 + 11) =
7x² + 4y² + 19 → sum
Then, we must subtract the sum to 4x² - 3y² - 7.
( 7x² + 4y² + 19) - (4x² - 3y² - 7) =
(7x² - 4x²) + (4y² + 3y²) + (19 + 7) =
Final Answer: 3x² + 7y² + 26
28.) -x + 2y + -2z
(-2x + y - 3z)+ ( x - y - z) =
(-2x + x) + (y + y) + (-3z + z) =
Final Answer: -x + 2y + -2z
29 and 30.)
Finding the perimeter of the rectangle:
Using the formula P = 2 (l + w), substitute the following given.
P = 2 ( 4b + 7) + (2b + 3)
P = 2 ( 6b + 10)
P = 12b + 20 m
Finding the perimeter of the square:
Using the formula P = 4 s, substitute the following given.
P = 4 (3b + 2)
P = 12b + 8 m
Therefore, perimeter of the rectangle is longer than the perimeter of the square.
12b + 20 - 12b + 8 = 12
The perimeter of the rectangle is 12m longer than the perimeter of the square.
