18 is my answer. You should still try to do the work yourself just to make sure its correct
Happy studying ^-^
B, because it’s saying 119 and over you can play on the junior varsity football team.
Answer:
If 
then 
and 
a | b | a + b (answer)
0 | 0 | 0
0 | 1 | 1
0 | 2 | 2
1 | 0 | 1
2 | 0 | 2
1 | 1 | 2
2 | 1 | 3
Step-by-step explanation:
Considering the following conditions for the real numbers:
Following the rules of these in-equations, it is possible to deduce:
Then, if the proposed statement is:
The conditions above shall comply the requirements established, but first, analyzing the statement:
If 
and 
then 
, 
and .
If and b a non negative real number, then , but because to , then 
. Due to the commutative property of sums, the same behavior will be presented if 
and a a non negative real number.
According to that, if , then and .
Answer:
oh ok thanks for free points
Step-by-step explanation:
Well, 50 points total given, 25 points per user and 13 bonus for brainliest
anyway
1.
add them equations, the y's will cancel
x+2y=4
<span>3x-2y=4 +</span>
4x+0y=8
4x=8
divide by 4 both sides
x=2
sub back
x+2y=4
2+2y=4
minus 2
2y=2
divide 2
y=1
x=2
y=1
(x,y)
(2,1) is solution
2.
the solution is where they intersect
multiply 2nd equation by 2 and add to first
4x-14y=6
<span>-4x+14y=-6 +</span>
0x+0y=0
0=0
infinite solutions
that is because they are actually the same line
the solutions are (x,y) such that they satisfy -2x+7y=-3 or 4x-14y=6 (same equaiton)
infinite solutions
3.
multiply first equation by 2 and add to first
4x+2y=-6
<span>1x-2y=-4 +</span>
5x+0y=-10
5x=-10
divide by 5 both sides
x=-2
sub bac
x-2y=-4
-2-2y=-4
add 2
-2y=-2
divide by -2
y=1
x=-2
y=1
(x,y)
(-2,1)
4.
coincident means they are the same line
so
we see that we have to multiply 4 by 2 to get 8
multiply top equation by 2
8x+10y=16
8x+By=C
B=10 and C=16
5.
a. false, either 0, 1, or infinity solutions
change the word 'two' to 'one' or 'zero' or 'infinite', or change 'can' into 'can't'
b.false
'sometimes' to 'always'
c. true
d. false, change 'sometimes' to 'always'
EC
total cost=150
TC=childC+adultC
TC=3c+5a
150=3c+5a
40 tickets, c+a
40=c+a
the equations are
150=3c+5a and
40=c+a
eliminate
multiply 2nd equaton by -3 and ad to first one
150=3c+5a
<span>-120=-3c-3a +</span>
30=0c+2a
30=2a
divide by 2
15=a
sub back
40=c+a
40=c+15
minus 15
25=c
25 children tickets and 15 adult tickets were sold
ANSWERS:
1.
(2,1) is solution
2.
infinite solutions
3.
(-2,1)
4.
B=10 and C=16
5.
a. false,
change the word 'two' to 'one' or 'zero' or 'infinite', or change 'can' into 'can't'
b.false
'sometimes' to 'always'
c. true
d. false, change 'sometimes' to 'always'
EC
the equations are
150=3c+5a and
40=c+a
25 children tickets and 15 adult tickets were sold
