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Sati [7]
3 years ago
6

Car lot one has a car to truck ratio of 3:4. Car lot two has a car to truck ratio of 2:3. Which car lot has more cars when both

lots have 12 trucks? How many cars would be on each lot? I need help asap please help!
Mathematics
1 answer:
Pani-rosa [81]3 years ago
4 0

Answer:

this means that lot 1 would have more cars. there were 3 cars in lot 1 and 2 in lot 2

Step-by-step explanation:

lot 1-          lot 2-

3:4             2:3

if both have 12 trucks, you need to multiply the number of cars with the number you used to make the trucks 12

3:4 would turn into 9:12

2:3 would be 8:12

this means that lot 1 would have more cars. there were 3 cars in lot 1 and 2 in lot 2

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Round 471.42857 to the nearest cent
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6 0
3 years ago
50 Points!
Advocard [28]
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
7 0
3 years ago
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