We can write this in math as x+y+z=104, x=y-6, and z=3y
Because we already know what x and z are in terms of y, we can substitute our values for x and z into the first equation. This now looks like (y-6) + y + (3y) = 104. Now we can simplify our equation to find our value for y.
y-6 + y + 3y = 104 simplifies to 5y - 6 = 104, then 5y=110, and finally y=22.
Now that we know our value for y we can find our values for x and z by substituting our value for y into the other two equations.
The second equation x = y-6 can be simplified as x = 22 - 6 and further simplified as x = 16.
The third equation z = 3y can be written as z = 3(22) or z = 66.
Our three numbers are 16, 22, and 66. Hope this helps you!
Look at it on the Internet
Answer— 45/15=3 so there is 3 classes with 15 students in each class? I didn’t really understand your question but I tried!
Hope it helped!
Given :
C, D, and E are col-linear, CE = 15.8 centimetres, and DE= 3.5 centimetres.
To Find :
Two possible lengths for CD.
Solution :
Their are two cases :
1)
When D is in between C and E .
. . .
C D E
Here, CD = CE - DE
CD = 15.8 - 3.5 cm
CD = 12.3 cm
2)
When E is in between D and C.
. . .
D E C
Here, CD = CE + DE
CD = 15.8 + 3.5 cm
CD = 19.3 cm
Hence, this is the required solution.
A: 1:1 unless they want you to count both arms then it would be B: 2:2
