12. How would you separate sand and water?
1 answer:
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My picture for 11 shows how you would set that up. S is the midpoint meaning it is directly in the center so both sides will be equal.
5x+17=8x-31
48=3x
16=x
now put that in for the x in RS
5(16)+17
80+17
RS=97
12)
same as 11 but you have a picture already. if a line bisects it then it is cut into 2 equal parts.
4-5x=2x+25
-21=7x
-3=x
put it in for both halves of AC
(4-5(-3))+(2(-3)+25)
4+15-6+25
AC=38
13)
You can find DE by subtracting CD by CE (49)
since we already have CE we just have to find CD which is equal to AC which has been set up in the same way as 11 and 12 except now we'll have to set up the equation differently.
2(3x+4)=11x-17
it's set up like this because we only have one half and the whole line.
2 times the half equals the lines length.
6x+8=11x-17
25=5x
5=x
put that in for the line AC
11(5)-17
55-17
38
and so CD equals 38
going back to the original solution:
DE=CE-CD
DE=49-38
DE=11
5x+17=8x-31
48=3x
16=x
now put that in for the x in RS
5(16)+17
80+17
RS=97
12)
same as 11 but you have a picture already. if a line bisects it then it is cut into 2 equal parts.
4-5x=2x+25
-21=7x
-3=x
put it in for both halves of AC
(4-5(-3))+(2(-3)+25)
4+15-6+25
AC=38
13)
You can find DE by subtracting CD by CE (49)
since we already have CE we just have to find CD which is equal to AC which has been set up in the same way as 11 and 12 except now we'll have to set up the equation differently.
2(3x+4)=11x-17
it's set up like this because we only have one half and the whole line.
2 times the half equals the lines length.
6x+8=11x-17
25=5x
5=x
put that in for the line AC
11(5)-17
55-17
38
and so CD equals 38
going back to the original solution:
DE=CE-CD
DE=49-38
DE=11