Answer:
11
Step-by-step explanation:
9
+ 2
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11
1 2 3 4 5 6 7 8 9 10 11
what the- to easy bro
Solve for m:3 m + 7/2 = 5/2 - 2 m
Put each term in 3 m + 7/2 over the common denominator 2: 3 m + 7/2 = (6 m)/2 + 7/2:(6 m)/2 + 7/2 = 5/2 - 2 m
(6 m)/2 + 7/2 = (6 m + 7)/2:(6 m + 7)/2 = 5/2 - 2 m
Put each term in 5/2 - 2 m over the common denominator 2: 5/2 - 2 m = 5/2 - (4 m)/2:(6 m + 7)/2 = 5/2 - (4 m)/2
5/2 - (4 m)/2 = (5 - 4 m)/2:(6 m + 7)/2 = (5 - 4 m)/2
Multiply both sides by 2:6 m + 7 = 5 - 4 m
Add 4 m to both sides:6 m + 4 m + 7 = (4 m - 4 m) + 5
4 m - 4 m = 0:6 m + 4 m + 7 = 5
6 m + 4 m = 10 m:10 m + 7 = 5
Subtract 7 from both sides:10 m + (7 - 7) = 5 - 7
7 - 7 = 0:10 m = 5 - 7
5 - 7 = -2:10 m = -2
Divide both sides of 10 m = -2 by 10:(10 m)/10 = (-2)/10
10/10 = 1:m = (-2)/10
The gcd of -2 and 10 is 2, so (-2)/10 = (2 (-1))/(2×5) = 2/2×(-1)/5 = (-1)/5:Answer: m = (-1)/5
Answer:
![y\in (-\infty ,5]](https://tex.z-dn.net/?f=y%5Cin%20%28-%5Cinfty%20%2C5%5D)
Step-by-step explanation:
In this problem, we need to write "Twenty times y is at most 100 in interval notation
".
20 times y means, 20y
Atmost means an inequality which is 
ATQ,
i.e.
We can also write it as :
Hence, the required interval notation is .
