The stopping distance of a car traveling 25mph is 61.7 feet, and for a car traveling 35 mph it is 106 feet. The stopping distance in feet can be described by the equation y = ax² +bx, where x is the speed in mph.
(a) Find the values of a and b.
(b)Use your answers from part (a) to find the stopping distance for a car traveling 55 mph.
y=ax^2+b
61.7=a*25^2 +b and
106=a*35^2 +b
subtracting the first from the second equation.
106-61.7=a(35^2-25^2) solve for a, then put that a into either equation to solve for b.
What are the features of a matrix?
What are the features of a matrix?
how do you solve this you have club a and b added together they equal 121 well if club a was increased by 4 times and club b was decresed by half what is a and b equal present
how do you solve this you have club a and b added together they equal 121 well if club a was increased by 4 times and club b was decresed by half what is a and b equal present
To solve this problem, we need to set up a system of equations using the given information.
Let's denote club A as "a" and club B as "b". We know that "a + b = 121" because the two clubs' totals added together equal 121.
We are also given the conditions that club A was increased by 4 times and club B was decreased by half. This can be expressed as "4a" for club A and "0.5b" for club B.
Now we can set up our equation:
4a + 0.5b = a + b
Simplifying this equation, we get:
3a - 0.5b = 0
To solve this equation, we can multiply both sides by 2 to eliminate the decimal:
6a - b = 0
Now we have a system of equations:
a + b = 121
6a - b = 0
We can solve this system by either substitution or elimination. Let's use the elimination method:
Multiply the first equation by -1, then add the two equations:
-a - b = -121
6a - b = 0
-5a = -121
Divide both sides by -5:
a = 24.2
Now we can substitute this value back into either of the original equations to solve for b:
24.2 + b = 121
b = 96.8
Therefore, club A is equal to 24.2 and club B is equal to 96.8 in the present.