1) The minimum and maximum number of items that he can purchase is
Given:
C=3x+5
X= (c-5)/3
Min:
X= (50-5)/3
X=15 items minimum
Max:
x = (80 - 5)/3 x = 25 items maximum 2) Find the maximum horizontal distance that the projectile may cover.
Using the quadratic formula
Maximum distance of projectile
3) The maximum height of the arrow is:
feet
Answer is 629 Feet
4) Find the number of customers that arrived in the 6th hour.
f(6) = 2^7 = 128
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64 customers arrived in the 6th hr.
5) If the net profit is 3800000, find the number of items sold.
P(x) =x^2-4000x + 7800000
3800000=x^2-4000x + 7800000
X^2-4000x+4000000=0
(x-2000) ^2=0
x=2000
P (2000) =3800000
Number of items sold=2000
6) Find its value after 3 years
f(x) = 20000(1/2)^x
20,000(.5)^3
20,000*.125=2,500 ans.
The value of the Machine after 3 years is $2,500.00
7) Find the maximum height that the object may reach.
f(x)=-x2+3x+6 use -b/2a
-3/2(-1)=-3/-2=3/2
x=3/2 f(3/2)=-(3/2)2+(3)(3/2)+6 f(3/2)=-(9/4)+(9/2)+6 f(3/2)=-(9/4)+(18/4)+6 f(3/2)=(9/4)+6 f(3/2)=(2 1/4)+6 f(3/2)=8 1/4 meters or 8.25 meters 8) The sum of two numbers is 210. If one number is the square of the other, find the set of numbers.
x +y = 210
y = x^2
substitute for y in first equation
x^2 +x -210 = 0
factor equation
(x+15)*(x-14) = 0
x = -15 or 14
y = 225 or 196
solution1 is x = -15, y = 225
solution2 is x = 14, y = 196
9) The area of a rectangle is 12 square inches. The length is 5 more than twice the width. Find the length and the width.
Area = 12 in2 = length * width The length is 5 more than twice the width: length = 5 + 2*width 12 = length*width
12 = (5 + 2*width)*width
12 = 5width + 2width2
0 = 2width2 + 5width - 12 Factors to: 0 = (2width-3)(width+4) width = 3/2 and -4 Since we can't have a negative length:
Width = 3/2 = 1.5 in.
Length = 5 + 2width = 5 + 2*1.5 = 8 in
10) The area of a rectangle is 28 square inches. The length is 8 more than thrice the width. Find the length and the width.
1) A=LW = 28 in
2) L = 3W + 8 Substitute L = 28/W into the 2nd equation 28/W = 3W + 8
28 = 3W^2 + 8W
3W^2 + 8W - 28 = 0
(3W+ 14)(W - 2) = 0 The answer must be positive therefore the second 'zero' is the answer W=2 inches
L=14 inches
