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Hello everyone, I have 22 questions for discrete mathematics course and I need a good expert who can helps me to solve them and its easy to solve them.
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Explanation & Answer
Please find attached the solution of all problems in word and PDF format. Please ask me if you have any doubt.Please note that Q-13 solution has been slightly changed from last solution. (Final answer is same, I have elaborated solution little bit more). Attached files have updated solution for that problem.Thanks.
1.
You have agreed to answer this.
2.
Equation of line: 𝑥 − 𝑦 = 2
Equation of parabola: 2𝑦 = 𝑥 2
To find intersection point:
From Equation of line, 𝑥 = 𝑦 + 2
Putting value of 𝑥 in Equation of parabola,
2𝑦 = (𝑦 + 2)2
⇒ 2𝑦 = 𝑦 2 + 4𝑦 + 4
⇒ 𝑦 2 + 2𝑦 + 4 = 0
⇒ (𝑦 2 + 2𝑦 + 1) + 3 = 0
⇒ (𝑦 + 1)2 = −3
This is not possible for any value of 𝑦, because (𝑦 + 1)2 ≥ 0∀𝑦 ∈ 𝑅
So, there is no point of intersection point between 𝒙 − 𝒚 = 𝟐 and 𝟐𝒚 = 𝒙𝟐 .
3.
Let, there were x students in the original group,
As, total cost = $1700
So, cost per student in original group =
total cost
number of students
=
$1700
x
As, 6 more students join the trip, number of students in new group = x + 6
As, for new group, cost per student drops by $7.50, so
1700
cost per student in new group =
− 7.5
x
As, for new group also, total cost should be $1700,
So,
cost per student in new group ∗ number of students in new group = 1700
1700
⇒(
− 7.5)(x + 6) = 1700
x
(1700 − 7.5x)(x + 6)
⇒
= 1700
x
⇒ (1700 − 7.5x)(x + 6) = 1700x
⇒ 1700x + 10200 − 7.5x 2 − 45x = 1700x
⇒ −7.5x 2 − 45x + 10200 = 1700x − 1700x
⇒ −7.5x 2 − 45x + 10200 = 0
⇒ −7.5(x 2 + 6x − 1360) = 0
⇒ x 2 + 6x − 1360 = 0
(Dividing both sides by -7.5)
⇒ x 2 + 40x − 34x − 1360 = 0
⇒ x(x + 40) − 34(x + 40) = 0
⇒ (x − 34)(x + 40) = 0
⇒ x = 34 or x = −40
As, number of students in the original group can’t be negative, so x = 34
So, there were 𝟑𝟒 students in the original group.
1700
1700
− 7.5 =
− 7.5 = 50 − 7.5 = 42.5
x
34
so, cost per student in new group is $ 𝟒𝟐. 𝟓
cost per student in new group =
4.
Let a, b and c are the short leg, long leg and hypotenuse of triangle respectively.
Here, 𝑏 = 𝑎 + 1
… (1)
also, perimeter = 12
⇒ 𝑎 + 𝑏 + 𝑐 = 12
⇒ 𝑐 = 12 − (𝑎 + 𝑏)
Putting 𝑏 = 𝑎 + 1 in above equation gives:
𝑐 = 12 − (𝑎 + 𝑎 + 1) = 12 − 2𝑎 − 1
⇒ 𝑐 = 11 − 2𝑎 … (2)
From Pythagoras Theorem:
𝑐 2 = 𝑎2 + 𝑏 2
Substituting 𝑐 and 𝑏 values from equation (1) and (2),
(11 − 2𝑎)2 = 𝑎2 + (𝑎 + 1)2
⇒ 121 − 44𝑎 + 4𝑎2 = 𝑎2 + 𝑎2 + 2𝑎 + 1
⇒ 121 − 44𝑎 + 4𝑎2 − 𝑎2 − 𝑎2 − 2𝑎 − 1 = 0
⇒ 2𝑎2 − 46𝑎 + 120 = 0
⇒ 2(𝑎2 − 23𝑎 + 60) = 0
⇒ 𝑎2 − 23𝑎 + 60 = 0(Dividing both sides by 2)
⇒ 𝑎2 − 3𝑎 − 20𝑎 + 60 = 0
⇒ 𝑎(𝑎 − 3) − 20(𝑎 − 3) = 0
⇒ (𝑎 − 20)(𝑎 − 3) = 0
⇒ a = 20 or a = 3
a = 20 is not possible, as side of a triangle can’t be greater than its perimeter.
So,
a=3
from, equation (1): 𝑏 = 𝑎 + 1 ⇒ b = 3 + 1 = 4
from, equation (2): 𝑐 = 11 − 2𝑎 ⇒ c = 11 − 6 = 5
so, length of each boundary of the land is: 3 miles, 4 miles and 5 miles.
5.
Let, I solved x problems correctly,
So, amount paid to me for correct solution = 8x
As, total problems solved are 26
So, number of problems solved incorrectly = (26 − x)
So, amount fined for incorrect solution = (26 − x) ∗ 5
As, we do not owe each other money
So, amount fined for incorrect solution = amount paid to me for correct solution
⇒ (26 − x) ∗ 5 = 8x
⇒ 130 − 5x = 8x
⇒ 130 = 13x
130
⇒x=
= 10
13
So, I solved 𝟏𝟎 problems correctly.
6.
If mechanic will be working on the job for x hours,
so mechanic charges for x hours = $34x
so, total cost = part cost + mechanic charges = 175 + 34x
minimum cost = $226
so, if xmin is minimum number of hours,
minimum cost = 175 + 34xmin
⇒ 226 = 175 + 34xmin
⇒ 226 − 175 = 34xmin
⇒ 51 = 34xmin
51
⇒
= xmin
34
⇒ xmin = 1.5 hours
maximum cost = $294
so, if xmax is maximum number of hours,
maximum cost = 175 + 34xmax
⇒ 294 = 175 + 34xmax
⇒ 294 − 175 = 34xmax
⇒ 119 = 34xmax
119
⇒
= xmax
34
⇒ xmax = 3.5 hours
So, the mechanic will be working on the job for 𝟏. 𝟓 − 𝟑. 𝟓 𝐡𝐨𝐮𝐫𝐬.
7.
As for each $1 increase in the ticket price, attendance decreases by 500
So, if there is $y increase in the ticket price,
Attendance will decrease by 500y.
a) ticket price after $y increase = present ticket price + y = 20+y
⇒ x = 20 + y
… (1)
Attendance after $y increase in the ticket price = present attendance – 500y = 30000 − 500y
⇒ N = 30000 − 500y
… (2)
From equation (1):
y = x − 20
Putting y’s value in equation (2):
N = 30000 − 500(x − 20)
⇒ N = 30000 − 500x + 10000
⇒ N = 40000 − 500x
So, the number of spectators at a football game, N, as a function of the ticket price, x:
𝐍 = 𝟒𝟎𝟎𝟎𝟎 − 𝟓𝟎𝟎𝐱
b) Revenue = ticket price*attendance = N*x
⇒ R = (40000 − 500x)x
⇒ R = 40000x − 500x 2
So, revenue from a football game, R, as a function of the ticket price, x:
𝐑 = 𝟒𝟎𝟎𝟎𝟎𝐱 − 𝟓𝟎𝟎𝐱 𝟐
8.
Let, x is the length of warehouse’s rectangular floor and y is the width of warehouse’s rectangular floor.
Area = Lenth ∗ width
⇒ xy = 4000
4000
⇒y=
… (1)
x
interior wall length = x
so, cost of interior wall = 125x
exterior wall length = perimeter of rectangular floor = 2(x + y)
so, cost of exterior wall = 175 ∗ 2(x + y) = 350(x + y)
so, contractor’s cost for building the walls = cost of interior wall + cost of exterior wall
⇒ C = 125x + 350(x + y)
Putting y =
4000
x
in above equation:
C = 125x + 350(x +
4000
x
)
1400000
x
𝟏𝟒𝟎𝟎𝟎𝟎𝟎
⇒ 𝐂 = 𝟒𝟕𝟓𝐱 +
𝐱
⇒ C = 125x + 350x +
9.
Let, the radius of inner semicircular ends is 𝑥 meters, so the radius of outer semicircular ends is 𝑥 + 4
meters.
So, length and width of outer rectangular portion are respectively: 180 meters and 2(𝑥+4) meters
So, area of outer recta...