<u>Step</u><u> </u><u>1</u>
given
<u>Step</u><u> </u><u>2</u>
<u>Step</u><u> </u><u>3</u>
Reason: Reflexive property
<u>Step</u><u> </u><u>4</u>
ASA
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.
75% of of the number 124 is 93
Answer:
The solution is (3/8, -7/8).
Step-by-step explanation:
y = −5x + 1
y = 3x − 2
Since the 2 expressions in x are both equal to y :
−5x + 1 = 3x - 2
-5x - 3x = -2 - 1
-8x = -3
x = 3/8.
So y = 3x - 2
= 3(3/8) - 2
= 9/8 - 2 = -7/8.
