MATH 1201 College Algebra Written Assignment Unit 1
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University of the People *
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Mathematics
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May 21, 2024
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University of the People
MATH 1201: College Algebra
Instructor Clifford Kettemborough
Written Assignment – Unit 1
November 22, 2023
Question 1: Determine Line Relationships
a. Equations:
{
3
y
+
4
x
=
12
;
-6y = 8x + 1
To determine if the lines are parallel, perpendicular, or neither, let's first rewrite the equations
in slope-intercept for
(
y
=
mx
+
b
)
First equation
y
=
−
4
3
x
+
4
Second equation y
=
−
4
3
x
−
1
6
The slopes are equal (-4/3), so the lines are parallel
b. Equations:
3
y
+
x
=
12
−
y
=
8
x
+
1
Solution:
Rewrite the equations in slope-intercept form.
First equation: y
=−
x
+
4
Second equation: y
=−
8
x
−
1
The slopes are not equal, so the lines are neither parallel nor perpendicular
.
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c. Equations 4
x
−
7
y
=
10
7
x
+
4
y
=
1
Solution: Rewrite the equations in slope-intercept form.
First equation: y
=
4
7
x
−
10
7
Second equation: : y
=
7
4
x
+
1
4
The slopes are negative reciprocals, so the lines are perpendicular.
Question 2: Ball Thrown in the Air
Height Function:
h
(
t
)
=−
4.9
t
2
+
24
t
+
8
a.
Height of the Building:
The height of the building is the initial value of the function, which is the constant term. Therefore, the height of the building is 88 meters.
h
(
0
)
=−
4.9
(
0
)
2
+
24
(
0
)
+
8
h
(
0
)
=
8
meter s
b.
Maximum Height:
The maximum height is the vertex of the parabolic function. The formula for the x-coordinate of the vertex is −
b
2
a
Plug
∈
a
=−
4.9
a
=−
4.9
∧
b
=
24
¿
findt .
t
=
−
24
(
2
) (
−
4.9
)
≈
2.45
seconds
Now, substitute t =2.45 into h(t) to find the maximum height
.
h(2.45)≈34.28 meters
c. Time to Reach Maximum Height:
The time to reach the maximum height is
t = 2.45 seconds
.
Question 3: Optimal Number of Shops
According to the question, the per day average income of a shop is $200 when 100
shops are in the market
.
Therefore, calculate the per day total income of the revenue department for 100 shops
as shown below
.
R
100=100×200=20,000
Therefore, the per day total income of the revenue department is $20,000
.
If 1 shop is added, the total number of shops is 101 in the market. Therefore, the per
day average income of a shop is decreased by $5. So, the per day average income of a
shop is $195
.
Calculate the per day total income of the revenue department for 101 shops as shown
below
.
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R
101=101×195=19,695
Therefore, the per day total income of the revenue department is $19,695
.
If 1 shop is decreased, the total number of shops is 99 in the market. Therefore, the
per day average income of a shop is increased by $5. So, the per day average income
of a shop is $205
.
Calculate the per day total income of the revenue department for 99 shops as shown
below
.
R
99=99×205=20,295
Therefore, the per day total income of the revenue department is $20,295
.
Now, on comparing the above calculations, it is observed that the per day total income
of the revenue department for 99 shops is more than the per day total income of the
revenue department for 100 and 101 shops
.
Hence, 99 shops should be in the market
.
To find the optimal number of shops for maximum total income (revenue) in the market,
we can use a quadratic equation
.
Let's assume the number of shops is x. The average income per shop will be (200 -
5x) if one more shop is added, and (200 + 5x) if one shop is removed
.
To maximize the total income, we need to find the value of x that maximizes the
product of the number of shops and the average income per shop
.
Total Income = x * (200 - 5x)
To find the optimal number of shops, we need to find the maximum value of this
quadratic equation. This can be done by finding the vertex of the quadratic function
.
The x-coordinate of the vertex can be found using the formula -b/2a, where a = -5 and
b = 200
:
x_vertex = -200 / (2 * -5)
x_vertex = 20
Therefore, the optimal number of shops for maximum total income is 20
.
Please note that this solution assumes linearity and does not consider other factors that
may affect the market's revenue. It is a simplified mathematical approach
.
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