<u>1/3; 2/7; 2/9</u>
Lets find least common multiple
The least common multiple is 63
1/3 --> 21/63
2/7 --> 18/63
2/9 --> 14/63
As you can see I am correct, but I found the LCM by multiplying 1/3 by 21 to get to 63 and the numerator as well
Answer:
N = 920(1+0.03)^4t
Step-by-step explanation:
According to the given statement a car repair center services 920 cars in 2012. The number of cars serviced increases quarterly at a rate of 12% per year after 2012.
Rate is 12 % annually
rate in quarterly = 12/4= 3%
We will apply the compound interest equation:
N=P( 1+r/n)^nt
N= ending number of cars serviced.
P= the number of cars serviced in 2012,
r = interest rate
n = the number of compoundings per year
t= total number of years.
Number of compoundings for t years = n*t = 4t
Initial number of cars serviced=920
The quarterly rate of growth = n=4
r = 3%
The growth rate = 1.03
Compound period multiplied by number of years = 920(1.03)^4t
Thus N = 920(1+0.03)^4t
N = number of cars serviced after t years...
Answer:
RS=5 TRUE
RS=4 FALSE (it’s 5)
ST=10 FALSE (it’s 8)
QR=4 FALSE (its 3)
Step-by-step explanation:
First, we have to determine the sides of the triangle.
4x-3=2x+1
4x-2x=1+3
2x=4
x=2
SO, RS=4x-3=4(2)-3=8-3=5
RT=2(x)+1=2(2)+1=5
QR=4²+b²=5². 16+b²=25. b²=25-16. b²=9. b=3
CLAIMS:
RS=5 TRUE
RS=4 FALSE (it’s 5)
ST=10 FALSE (it’s 8)
QR=4 FALSE (its 3)
The distance between Car A and car B,When Car A crosses the start line is:
distance =speed car B* time
distance=(15 m/s)(3 s)=45 m
Distance traveled by car A =x, (when the car B is at the same distance from the start line)
time of car A=t
x=10 m/st ⇒ x=10t (1)
Distance traveled by car B=x
time of car B=t-3
x=15(t-3) (2)
With the equations (1) and (2) we make a system of equations:
x=10t
x=15(t-3)
We solve this system of equations:
10t=15(t-3)
10t=15t-45
-5t=-45
t=-45 / -5
t=9
t-3=9-3=6
x=10 t=10 (9)=90
Answer: The time would be 9 seconds for Car A and 6 seconds for car B and the distance would be 90 meters.
Assuming Earth's gravity, the formula for the flight of the particle is:
s(t) = -16t^2 + vt + s = -16t^2 + 144t + 160.
This has a maximum when t = -b/(2a) = -144/[2(-16)] = -144/(-32) = 9/2.
Therefore, the maximum height is s(9/2) = -16(9/2)^2 + 144(9/2) + 160 = 484 feet.
