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.
Ok so to find the eggs left we need to subtract but with fractions, you need a common denominator. You can multiply 1/6 by 2 because 6·2=12 which will be the same denominator as 7/12. You must multiply the numerator by 2 as well so 1/6· 2/2= 2/12. Now subtract 7/12 - 2/12 and you have 5/12 eggs left. :)
The first option :
Multiply the first equation by 8 and the second equation by 2
would not result in the elimination of the variables in this system
Answer: it's just a black screan
Step-by-step explanation:
Answer:
3n - 4m
Step-by-step explanation:
(2n-5m)+(n+m)
so we get this equation and from here, we expand the equation by removing the brackets, you should get:
2n-5m+n+m and just solve from here.
to make it easier, put the like terms together:
2n+n+m-5m
= 3n-4m