Answer:6.042km/h 6km/h approximately
Step-by-step explanation:
First off we have to know the formula relating speed, distance and time which is
Speed = distance/time
Now we are looking for Rachel's running speed
We are to find Rachel's running speed, so let's label is x
We are given that the distance Rachel runs to her bus stop is 2km
We were not given the time she uses to run to the bus stop
So let's label the time Rachel uses to run to her bus stop as y
So from the formula speed = distance/time
We have x = 2/y
Now we are told that the speed the bus uses to get to school is 45km/h faster than her speed used to run
So speed of bus = 45 + x
And the overall time for the whole journey is 25mins, changing this to hours, because the speed details given is in km/h we divide 25 by 60 which will give 0.417
Now if the total time is 0.417 hours, and we labeled the time for Rachel to run to the bus as y, so the time for the time for the bus to get to school will be 0.417 - y
We are also told the bus rides for 4.5km to school
So adding this together to relate the speed, distance and time of the bus with the formula speed = distance/time
We get 45 + x = 4.5/(0.417 - y)
So we have two equations
x = 2/y (1)
45+x = 4.5/(0.417-y) (2)
So putting (1) in (2) we have
45 + (2/y) = 4.5/(0.417-y)
Expanding further
(45y + 2)/y = 4.5(0.417-y)
Cross multiplying
(45y + 2)(0.417 - y) = 4.5y
Opening the brackets
18.765y - 45y2 + 0.834 - 2y = 4.5y
Collecting like terms
-45y2 + 18.765y -2y - 4.5y + 0.834 = 0
-45y2 + 12.265 + 0.834 = 0
Dividing all sides by -45 to make the coefficient of y2 1
y2 - 0.273y - 0.019 = 0
Now we have gotten a quadratic equation, and since it's with decimal numbers we can use either completing the square method of almighty formula
I'm using almighty formula her
For solving
ax2 + bx + c = 0
x = (-b +-root(b2-4ac)/2a
For our own equation, we are finding y
From our our quadratic equation
a = 1, b=-0.273, c = -0.834
you = (-(-0.273)+-root(-0.273-4(1)(-0.019))/2(1)
y = (273+-root(0.151))/2
y = (0.273+0.389)/2 or (0.273-0.389)/2
y = 0.331 or -0.085
So we use the positive answer which is 0.331, because time can't be negative
Then we put y = 0.331 in (1)
x = 2/y
x = 2/0.331
x = 6.042km/h
x = 6km/h approximately
are all real numbers
we're translating the function horizontally, so the range remains the same
we're stretching the function horizontally, so the range doesn't change
we're translating the function 1 unit up, but the range is already all the real numbers, so it doesn't change.