For this case, the first thing we must do is define variables.
We have then:
n: number of cans that each student must bring
We know that:
The teacher will bring 5 cans
There are 20 students in the class
At least 105 cans must be brought, but no more than 205 cans
Therefore the inequation of the problem is given by:
Answer:
105 <u><</u> 20n + 5 <u><</u> 205
the possible numbers n of cans that each student should bring in is:
105 <u><</u> 20n + 5 <u><</u> 205
It always helps to draw a picture. Given the information, Segment OF is the center line that is bisecting this angle.
Since it's bisecting (cutting in half)... we can simply set the two angles equal to each other.
y+30=3y-50
-2y+30=-50
-2y=-80
y = 40
C)40.
Answer: The answer would be 8.
Step-by-step explanation: I'm going to assume you mean 7 8/13 is a mixed fraction. So you multiply the denominator, 13, multiply 7 to that, then add 8 to it and the denominator would be the same as before. That should give you 99/13. When you divide, you get around 7.6, rounding closest to 8.
