Problem 1:
1) 3/5x-1 x 3/4=9/10
3x/5-1 x 3/4=9/10
2) 3x/5-1 x 3/4=9/10
3x/5-3/4=9/10
3) 3x/5-3/4=9/10
+3/4 +3/4
4) 3x/5=33/20
5) 5 x 3x/5= 5 x (33/20)
6) 3x=33/4
7) 3x=33/4
/3 /3
8) x=11/4
Problem 2:
1) 6/7+2/5x=-4/5
6/7+2x/5=-4/5
2) 6/7+2x/5=-4/5
-6/7 -6/7
3) 2x/5= --58/35
4) 5 x 2x/5= 5 x (--58/35)
5) 2x= --58/7
6) 2x= --58/7
/2 /2
x= --29/7
The answer is a i hope i helped
Answer:
option 4.
16 square units
Step-by-step explanation:
as we do not have the measures of the sides, but if the points of the vertices with Pythagoras we can calculate the sides.
P = (2 , 4)
S = (4 , 2)
we have to subtract the values of p from s
PS = (4 - 2 , 2 - 4)
PS = (2 , -2)
by pitagoras h ^ 2 = c1 ^ 2 + c2 ^ 2
h: hypotenuse
c1: leg 1
c2: leg 2
PS^2 = 2^2 + -2^2
PS = √ 4 + 4
PS = √8
PS = 2√2
S = (4 , 2)
R = (8 , 6)
SR = (8-4 , 6-2)
SR = (4 , 4)
by pitagoras h ^ 2 = c1 ^ 2 + c2 ^ 2
h: hypotenuse
c1: leg 1
c2: leg 2
SR^2 = 4^2 + 4^2
SR = √ (16 + 16)
SR = √32
SR = 4√2
having the values of 2 of its sides we multiply them and obtain their area
PS * RS = Area
2√2 * 4√2 =
16
