Problem 1 :
The function f(t) = -5t^{2} + 20t + 60 models the approximate height of an object t seconds after it is launched. How many seconds does it take the object to hit the ground?
Solution:
f(t) = -5t^{2} + 20t + 60
= 5t^{2} - 20t - 60
Dividing by 5
t^{2} - 4t - 12 = 0
t^{2} - 6t + 2t - 12 = 0
t(t - 6) + 2(t - 6) = 0
(t + 2) (t - 6) = 0
t + 2 = 0 t = -2 |
t - 6 = 0 t = 6 |
As time cannot be negative, the object will hit the ground after 6 seconds.
Problem 2 :
What is the smallest of 3 consecutive positive integers if the product of the smaller two integers is 5 less than 5 times the largest integer?
Solution:
Let the smallest number be x, and the second and third be x + 1 and x + 2.
x(x + 1) = 5(x + 2) - 5
x^{2} + x = 5x + 10 - 5
x^{2} - 4x - 5 = 0
(x - 5) (x + 1) = 0
x = 5 and x = -1
The numbers have to be positive, the smallest number is 5.
x + 1 = 5 + 1 = 6
x + 2 = 5 + 2 = 7
So, the 3 consecutive integers are 5, 6 and 7.
Problem 3 :
The larger leg of a right triangle is 3 cm longer than its smaller leg. The hypotenuse is 6 cm longer than the smaller leg. How many centimeters long is the smaller leg?
Solution:
Let x represent the shorter leg.
Then the length of the longer will be x + 3 and
the length of the hypotenuse will be x + 6.
By using Pythagorean Theorem,
x^{2} + (x + 3)^{2} = (x + 6)^{2}
x^{2} + x^{2} + 6x + 9 = x^{2} + 12x + 36
2x^{2} - x^{2} + 6x - 12x + 9 - 36 = 0
x^{2} - 6x - 27 = 0
(x - 9) (x + 3) = 0
x = 9 or x = -3
The length can't be negative.
So, x = 9
The length of the shorter leg is 9 centimeters.
Problem 4 :
Which term is a factor of 3a^{2} + 12a?
A. 3a B. 4a C. 3a^{2 }D. 4a^{2}
Solution:
= 3a² + 12a
= 3a(a + 4)
So, option (A) is correct.
Problem 5 :
Which graph displays function f(x) = (2x + 3) (x - 2)?
Solution:
f(x) = (2x + 3)(x - 2)
x-intercept : put y = 0
2x + 3 = 0 2x = -3 x = -3/2 x = -1.5 |
x - 2 = 0 x = 2 |
Converting into vertex form :
f(x) = (2x + 3)(x - 2)
f(x) = 2x^{2} - 4x + 3x - 6
f(x) = 2x^{2} - x - 6
f(x) = 2[x^{2} - (1/2)x] - 6
f(x) = 2[x^{2} - (1/2)x] - 6
So, graph (a) is correct.
Problem 6 :
The floor of a rectangular cage has a length 4 feet greater than its width, w. James will increase both dimensions of the floor by 2 feet. Which equation represents the new area, N, of the floor of the cage?
A. N = w^{2} + 4w B. N = w^{2} + 6w C. w^{2} + 6w + 8
D. w^{2} + 8w + 12
Solution:
Given, the original width = w
The original length = w + 4
Both length and width will be increased by 2 feet
So, new length = w + 6, new width = w + 2
New area N = (w + 6) (w + 2)
= w^{2} + 2w + 6w + 12
N = w^{2} + 8w + 12
So, option (D) is correct.
Problem 7 :
Which expression is equivalent to t^{2} - 36?
A. (t - 6)(t - 6) B. (t + 6)(t - 6)
C. (t - 12)(t - 3) D. (t - 12)(t +3)
Solution:
= t^{2} - 36
= t^{2} - 6^{2}
(t + 6) (t - 6)
So, option (B) is correct.
Problem 8 :
Draw the graph of the function f(x) = 4x^{2} - 8x + 7?
Solution:
f(x) = 4x^{2} - 8x + 7
Let us find vertex and zeroes of the quadratic function.
f(x) = 4[x^{2} - 2x] + 7
f(x) = 4[(x - 1)^{2} - 1] + 7
= 4(x - 1)^{2} - 4 + 7
= 4(x - 1)^{2} + 3
y-intercept: x = 0
f(x) = 7
y-intercept = (0, 7)
x-intercept: y = 0
4(x - 1)^{2} + 3 = 0
Will get unreal value, so there is x-intercepts.
So, option (D) is correct.
Problem 9 :
Suppose that the equation V = 20.8x^{2} - 458.3x + 3,500 represents the value of a car from 1964 to 2002. What year did the car have the least value? (x = 0 in 1964)
A. 1965 B. 1970 C. 1975 D. 1980
Solution:
V = 20.8x^{2} - 458.3x + 3500
To find the least value, we can figure out the minimum value.
x = -b/2a
x = 458.3/2(20.8)
When V' = 0, it has least value
41.6x - 458.3 = 0
41.6x = 458.3
x = 458.3/41.6
x = 11.01
1964 + 11 = 1975
So, option (C) is correct.
Problem 10 :
The number of bacteria in a culture can be modeled by the function N(t) = 28t^{2} - 30t + 160, where t is the temperature, in degrees Celsius, the culture is being kept. A scientist wants to have fewer than 200 bacteria in a culture in order to test a medicine effectively. What is the approximate domain of temperatures that will keep the number of bacteria under 200?
A. -1.01°C < t < 2.03°C B. -0.90°C < t < 1.97°C
C. -0.86°C < t < 1.93°C D. -0.77°C < t < 1.85°C
Solution:
N(t) = 28t^{2} - 30t + 160
200 = 28t^{2} - 30t + 160
28t^{2} - 30t - 40 = 0
By using quadratic formula,
a = 28
b = -30
c = -40
-0.77°C < t < 1.85°C
So, option (D) is correct
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