3 quarters, and two thirds of a quarter so It's [3 2/3]
Speed
Riding speed = 9 mph
Driving speed = 27 mph
Time
If x hours is the time taken when riding, (x - 1/3) hours is the time taken when driving.
Distance in both instances is the same, D= Speed * time
Therefore,
9*x = 27(x-1/3)
9x = 27x - 27/2
9x = 27x - 9
(9-27)x = -9
-18x = -9
x = 9/18 = 1/2 hours
Distance, D = 1/2 *9 = 4.5 miles
Answer:
b
Step-by-step explanation:
you dont need a explanation do you
This can be solved in two ways: With heavy tools or with just algebra.
What is your level? Have you studied calculus?
With pure algebra:
We need to find the maximum of the function <span>h = −16t^2 + 36t + 5
Lets take out -1 for simplicity:
</span><span>h = −(16t2 - 36t - 5)
For now lets just work with this: </span>16t^2 - 36t - 5
16t^2=(4t)^2
(4t-x)^2= 16t^2-2*4xt+x^2
we have -36t so x should be 4.5 as 2*4*4.5=36
Lets see what we have now:
16t^2 - 36t - 5= (4t-4.5)^2 is this true? No but close
(4t-4.5)^2= 16t^2- 2*4*4.5t +4.5^2= 16t^2-36t+20.25
16t^2 - 36t - 5 and 16t^2-36t+20.25 nearl the same just take away 25.25 from the right hand side
Getting long, just stay with me:
16t^2 - 36t - 5= (4t-4.5)^2 - 25.25
h= -{(4t-4.5)^2 -25.25}
h=-(4t-4.5)^2 + 25.25
We want to find the maximum of this function. -(4t-4.5)^2 this bit is always negative or 0, so it maximum is when it is 0. Solve: 4t-4.5=0
t=1,125
Answer:
1) 90 mi
2) 4 inch
Step-by-step explanation:
1)
Given that the distance between Fairview and Abbots Rise on the map = 5 in
Scale of the map = 1 in: 18 mi
i.e. 1 inch on map is equivalent to an actual distance equal to 18 mi between the cities.
Given that the map has a distance of 5 inches.
Therefore, the actual distance between the cities = 5 
18 = <em>90 mi</em>
<em></em>
2) A particular satellite is 12 ft wide.
Model's scale = 1 in : 3 ft
i.e. 1 inch on map is equivalent to actual distance of 3 ft.
OR
3 ft of actual distance = 1 in on map
1 ft of actual distance = 
in on map
12 ft of actual distance = 
