BID Test
The formula h=-16t^2 + 32t + 80 gives the distance in feet above the ground, h, reached by an object in t seconds if thrown straight up from the top of an 80 ft building. A) what is the maximum height reached by an object? B) how long to reach max height? C) how long does it take for object to hit the ground?
A) h=-16t^2 + 32t + 80 h=-16(t^2 - 2t + 1) +80 + 16 h=-16(t-1)^2 +96 h=96 feet B) 96=-16(t-1)^2 + 96 0=-16(t-1)^2 0=t-1 t=1 sec C) 0=-16(t-1)^2 +96 -96=-16(t-1)^2 6=(t-1)^2 Rad(6)=t-1 Rad(6)+1= t t=3.45=3.5 sec
A wire 360 inches long is cut into two pieces. One piece makes a square and another a circle. If the figures have the same area, what are the lengths of the the two wires pieces (to nearest tenth of an inch)?
L=length of square R=radius of circle Circle area=(pi)r^2 Square area=4L (pi)r^2=L^2 L=square root of ((pi)r^2) L=square root of (pi)x radius 2(pi)r=circumference 4L=perimeter of square 4L + 2(pi)r =360 4(square root of (pi) x r) + 2(pi)r =360 4(square root of pi)r + 2(pi)r =360 r=360/(4(radical pi)) + 2(pi) r=26.92 in 2(pi)r=169.14 in 360-169.14=190.9 in Answer: (190.9 in., 169.1 in.)
Find the number of units that produce max revenue. R=400x-0.01x^2, where R is the total revenue in dollars and x is the number of units sold.
R=400x -0.01x^2 x=-b/2a -400/2(-0.01) x=20,000 units
Find the quadratic function whose vertex is (3,-1) and one x-intercept is 1.
V=(3,-1) X=1 y=a(x-h)^2 + k 0=a(1-3)^2 -1 0=a(-2)^2 -1 0=4a-1 a=1/4 y=1/4(x-3)^2-1
Find the equation of a given parabola that has a max value of 5 if the parabola is of the form y=ax^2+3x-4.
Vertex=(h,5) h=-b/2a 5=a(h)^2+3(h)-4 5=a(-3/2a)^2 +3(-3/2a) -4 5=a(9/4a^2) -9/2a -4 5=9/4a-9/2a-4 5=-9/4a-4 9=-9/4a 36a=-9 a=-1/4 h=-3/2(-1/4) h=-3/(-1/2) h=6 y=-1/4(x-6)^2 + 5
The marketing dept of the TENRAQ Tennis Company found that, on avg, 600 rackets will be sold monthly at the price of $100. The dept also observed that for each $5 reduction in price, an extra 50 rackets will be sold monthly. What price will bring in the largest monthly income?
p=100-5x I=yp I=(50x + 600)(100-5x) I=5000x+60,000-250x^2-3000x I=-250x^2 + 2000x +60,000 Vertex=(h,k) h=-b/2a h=-2000/2(-250) h=4 p=100-5(4) p=100-20 p=$80
A horizontal bridge is in the shape of a parabolic arch. Given the info shown in figure 16, what is the height, h, of the arch 2 feet from the shore?
y=ax^2 + bx + c y=ax^2 + bx + 10 Points= (-10,0), (10,0), (0,10) y=100a -10b +10 | y=100a+10b+10 Combine: 0= 200a + 20 -20=200a a=-0.1 y=-0.1x^2 + 10 y=-0.1(8)^2 + 10 y=-0.1(64) + 10 y=3.6 feet
