First picture)
I: 5x+2y=-4
II: -3x+2y=12
add I+(-1*II):
5x+2y-(-3x+2y)=-4-12
8x=-16
x=-2
insert x=-2 into I:
5*(-2)+2y=-4
-10+2y=-4
2y=6
y=3
(-2,3)
question 6)
I: totalcost=115=3*childs+5*adults
II: 33=adults+childs
33-adults=childs
insert childs into I:
115=3*(33-adults)+5*adults
115=99-3*adults+5*adults
16=2*adults
8=adults
insert adults into II:
33-8=childs
25=childs
so it's the last option
question 7)
a) y<6 and y>2 can also be written as 2<y<6, so solution 3 exist for example
b) y>6 and y>2 can also be written as 2<6<y, so solution 7 exist for example
c) y<6 and y<2 inverse of b: y<2<6, so for example 1
d) y>6 and y<2: y<2<6<y, this is impossible as y can be only either bigger or smaller than 2 or 6
so it's the last option
question 8)
I: x+y=12
II: x-y=6
subtract: I-II:
x+y-(x-y)=12-6
2y=6
y=3
insert y into I:
x+3=12
x=9
(9,3)
question 9)
I: x+y=6
II: x=y+5
if you take the x=y+5 definition of II and substitute it into I:
(y+5)+y=6
which is the second option :)
The answer is ten plz thank
Answer:
DE=145
Step-by-step explanation:
Both circles are congruent and so are the triangles, because of the radius. So mDE is also 145
The answer would be B because y=mx+b
X=amount of money woman has
n=pounds of sugar
Store #1: 8.5*n=x+30
Store #2: 8*n=x-25
multiply #2 by 17/16 to make coefficient of n same as in #1
2: 8.5*n=17/16(x-25)
Subtract #2 from #1
0=x+30-17/16(x-25)
x+30-17x/16 +25*17/16
x/16=30+25*17/16
x=480+425
=905 cents
=$9.05
