X= -10.5+9.5
X=1
I hope that helped:)
Answer:
Step-by-step explanation:
This question can be solved by a system of equations.
I am going to say that:
x is the number of regular fares paid.
y is the number of reduced fares paid.
1,042 total fares were paid
This means that
A city bus collected $1,140 in fares on a given day. The regular fare is $1.50, and the reduced fare is $0.75.
This means that:
System of equations:
The system of equations is:
Answer:
Step-by-step explanation:
Hello,
At the beginning the population is 240,000
After 1 year the population will be
240,000*(1+7.75%)=240,000*1.0775
After n years the population will be
So after 7 years the population will be
So rounded to the nearest whole number gives 404,699
Hope this helps
ANSWER: ⇒ ¹/₃
Solve the equation:
6a - 4 = -2
Rearrange variables to the left side of the equation:
6a = -2 + 4
Calculate the sum or difference:
6a = 2
Divide both sides of the equation by the coefficient of variable:
a = ²/₆
Cross out the common factor:
a = ¹/₃
Answer:
C. Ari and Matthew collide at 4.8 seconds.
Explanation:
Ari and Matthew will collide when they have the same x and y position. Since Ari's path is given by
x(t) = 36 + (1/6)t
y(t) = 24 + (1/8)t
And Matthew's path is given by
x(t) = 32 + (1/4)t
y(t) = 18 + (1/4)t
We need to make x(t) equal for both, so we need to solve the following equation
Ari's x(t) = Matthew's x(t)
36 + (1/6)t = 32 + (1/4)t
Solving for t, we get
36 + (1/6)t - (1/6)t = 32 + (1/4)t - (1/6)t
36 = 32 + (1/12)t
36 - 32 = 32 + (1/12)t - 32
4 = (1/12)t
12(4) = 12(1/12)t
48 = t
It means that after 48 tenths of seconds, Ari and Mattew have the same x-position. To know if they have the same y-position, we need to replace t = 48 on both equations for y(t)
Ari's y position
y(t) = 24 + (1/8)t
y(t) = 24 + (1/8)(48)
y(t) = 24 + 6
y(t) = 30
Matthew's y position
y(t) = 18 + (1/4)t
y(t) = 18 + (1/4)(48)
y(t) = 18 + 12
y(t) = 30
Therefore, at 48 tenths of a second, Ari and Mattew have the same x and y position. So, the answer is
C. Ari and Matthew collide at 4.8 seconds.