There are 6 peers of whole numbers that have a sum of 110.
Answer:s>_18
Step-by-step explanation:
First distribute the 1/5 through the parentheses.
1/5 (3 + 2z) - 4z = -3
3/5 + 2/5z - 4z = -3
subtract 3/5 from each side
2/5z - 4z = -3 - 3/5
combine like terms
-3 3/5z = - 3 3/5
z = 1
Answer A
Answer:
The inequality to determine the number of rotations is given by
r ≤ 5.25
Therefore the maximum number of rotations possible is 5.
Step-by-step explanation:
i.) The bottom of Ignacio's desktop is 74.5 cm from the floor.
ii.) The tops of his legs are 49.3 cm from the floor.
iii.) the difference between the bottom of Ignacio's desktop and the top of his legs = 74.5 - 49.3 = 25.2 cm.
iv.) Each clockwise rotation of the knob on the chair raises Ignacio's legs by 4.8 cm.
v.) let the number of clockwise rotations be r.
vi.) therefore the inequality to determine the number of rotations is given by
4.8r ≤ 25.2 therefore r ≤ 25.2/4.8 therefore r ≤ 5.25