So I'm assuming that you're taking Calculus.
The first thing you want to do is take the integral of f(x)...
Use the power rule to get:
4X^2-13X+3.
Now solve for X when f(x)=0. This is because when the slope is 0, it is either a minimum or a maximum(I'm assuming you know this)
Now you get X=0.25 and X=3. Since we are working in the interval of (1,4), we can ignore 0.25
Thus our potential X values for max and min are X=1,X=4,X=3(You don't want to forget the ends of the bounds!)
Plugging these value in for f(x), we get
f(1)=2.833
f(3)=-8.5
f(4)=1.667
Thus X=1 is the max and X=3 is the min.
So max:(1,2.833)
min:(3,-8.5)
Hope this helps!
4 kilometres is 4000 meters
4000 meters - 3400 meters = 600 meters
and as John is the one that rode 4 kilometres he rode further and by 600 meters
Answer:
d
Step-by-step explanation:
BC supplementary means they add up to 180 and answer d adds up to 180
D. 75 cm
Step-by-step explanation:
fill in the y with 20.5, so you get 20.5=0.3x-2. Add the 2 to the 20.5 so you get 22.5=0.3x. Divide both by 0.3 to get 75=x.
sorry if it's a bd explanation, hope this helps :)
Answer:
6.667 ft/s
Step-by-step explanation:
Given
Height of street light = 15ft
Height of man = 6ft
Speed away from pole = 4ft/s
Let x represents the distance between the man and the pole, and Let y represents the distance between the tip of the man's shadow to the pole.
This forms a similar triangle (see attachment below).
From similar triangles, we have
(y - x)/6 = y/15 ----- Solve equation
15(y - x) = 6 * y
15y - 15x = 6y ---- Collect Like Terms
15y - 6y = 15x
9y = 15x --- divide through by 9
y = 15x/9
y = 5x/3 ---- differentiate with respect to time, t
dy/dt = 5/3 dx/dt
dx/dt represents rate of change of distance per time = 4ft/s
while dy/dt represents rate of movement of his shadow tips
dy/dt = 5/3 * 4
dy/dt = 20/3 = 6.667 ft/s