Answer:
Ignacio can make 5 clockwise rotations.
Step-by-step explanation:
Given that the Ignacio's legs are already at a height of 49.3cm and each rotation of the chair knob raises his legs another 4.8cm, we can set up an inequality to determine the number of rotations Ignacio could make without his legs touching the desk, which is at a height of 74.5cm:
4.8r + 49.3 < 74.5 where 'r' is equal to the number of rotations
The sum of the Ignacio's original leg height plus the amount of height increased from the rotations of the know must be less than 74.5 in order for his legs not to touch. Now, solve for 'r':
Subtract 49.3 from both sides: 4.8r + 49.3 - 49.3 < 74.5 - 49.3 or 4.8r < 25.2
Divide 4.8 from both sides: 4.8r/4.8 < 25.2/4.8 or r < 5.25
Since the number of rotations must be less than 5.25, he can make 5 complete rotations.
Let the second no. be x
-3x = 21 → x = 21/-3 → x = -7
2/5 * 10 = 20/50
0.22 as a fraction 11/50
20/50 is the same as 40/100
11/50 is the same as 22/100
multiply
880/10000
which is 22/250
<span>2a(7a²-3a+9)
=14a^3 - 6a^2 +18a
answer is </span><span>D. 14a³-6a²+18a</span>
Answer:
hi
Step-by-step explanation:
