Unit 4 Assessment
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Unit 4 Assessment: Solve Application Problems Factoring Polynomials Amoi Foster Herzing University MA109-7A College Algebra
Professor Patricia Manderville February 3
rd
, 2024
Question:
An integer is 3 less than 5 times another. If the product of the two integers is 36, then find the integers.
Answer: x * y= 36 y = 5x –3
x(5x – 3)=36
5x2 – 3x - 36 =0
(5x *12) (x -3) =0
x=3
y=12
3*5=15 and 15-3=12. The product of 3 and 12 is 36.
Question:
The width of a rectangle is 5 units less than the length. If the area is 150 square units, then find the dimensions of the rectangle.
Answer:
l= w+5
l*w=150 sq. Units
w(w + 5)=150
w^2 + 5w - 150 = 0
(w - 10) (w+ 15) = 0
w = -15, 10=> reject negative width:
w=10 units
l=15 units
Questions:
The length of a rectangle is 4 inches more than its width. The area of the rectangle is equal to 5 inches more than 2 times the perimeter. Find the length and width of the rectangle.
Answer: The length of the rectangle is 4 inches more than its width
Let l = length, and w = width
l=4+w ------ equation (1)
The area of the rectangle is equal to 5 inches more than 2 times the perimeter then
a= 5+2p --- equation (2)
Let the area be A and the perimeter be P
Area of rectangle is given by the formula
a=l x w = lw
And the perimeter of a rectangle is
p= 2(1+w)
Putting into equation (2)
a= 5+2p
We have
lw=5+2[2(l+w)]
lw=5+22(2l+2w)
lw=5+4l+4w --- equation (3)
From equation (1)
l=4+w
substitute this into equation (3)
lw=5+4l+4w
(4+w)w=5+4(4+=w)+4w
4w+w^2=5+16+4w+4w
w^2+4w=21+8w
w^2+4w-8w-21=0
w^2-4w-21=0
Solve equation quadratically
W^2-4w-21=0
W^2-7w+3w-21=0
W(w-7)+3(w-7)=0
(w+3)(w-7)=0
W+3=0 or w-7=0
W=-3 or w=7
Since dimension cannot be negative
Width = 7 inches
For length
Substitute the value of w into equation (1)
L=4+w
L=4+7
Length is 11 inches
The length of the rectangle is 11 inches and width is 7 inches.
Questions: The height of a projectile launched upward at a speed of 32 feet/second from a height
of 128 feet is given by the function. How long will it take h(t)= -16t^2+32t+128 projectile to hit the ground? Answer: h(t)=−16t2+32t+48
When t=0
h(0)=0+0+48ft
The projectile will hit the ground when h(t)=0 therefore, −16t^2+32t+48=0
16t^2−32t−48
16(t^2−2t−3) = 0
(t+1)(t−3) = 0
the time is = 3 seconds
Questions: 16t^2 5. The height of an object dropped from the top of a 144-foot building is given by (t) = +144. How long will it take the object to hit the ground?
ℎ
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Answer:
h(t) = -16t^2 + 144
Solve for t when h(t) = 0
0 = -16t^2+144
16t^2 = 144
t= √(144/16) = 3 sec
Proof:
0 = -16*(3)^2 + 144?
0 = -144 + 144?
t = 3 seconds