Answer:
If EFGH is an isosceles trapezoid,
if
EH=4x-27+
FG=x%2B9+
EG=3y%2B19
FH=11y-21
then EH=FG and EG=FH
set up equations:
4x-27=x%2B9 .......solve for x
4x-x=27%2B9
3x=36
x=36%2F3
3y%2B19=11y-21.......solve for y
11%2B19=11y-3y
30=8y
y=30%2F8
y=15%2F4
y=3.75
so,x=12 and y=3.75
you can also check the length of equal sides and equal diagonals:
EH=4%2A12-27+=21
FG=12%2B9+=21
EG=3%283.75%29%2B19=30.25
FH=11%283.75%29-21=30.25x=12
Step-by-step explanation:
Answer:
Don't know because there is only 1 number there is no another . please write answer in comment
Answer:
x=5
Step-by-step explanation:
Graph each side of the equation. The solution is the x-value of the point of intersection.
Answer:
72
Step-by-step explanation:
Jeff can divide the coin evenly in 6 or 9 bags, so the number of coin should have 9 and 6 as the factor. The lowest common factor for 6 and 9 will be:
6= 2 x 3
9= 3^2
LCF= 2 x 3^2= 18
From here you need to find the common multiple of 18 that will have 2 remainder after divided by 5. When 18( divided by 5, it will left 3 as remainder.
Multiple of 3 that give 2 remainder after divided by 5 will be 3x4= 12. Then least coin Jeff should have will be: 4x18= 72
