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6 pages/≈1650 words
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APA
Subject:
Mathematics & Economics
Type:
Math Problem
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English (U.K.)
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Computation of a Mathematical Equation (Math Problem Sample)

Instructions:

1:sOLVE x^2-2x-15=0 USING THREE METHODS OF SOLVING QUADRATIC EQUATION.
2:a>b then 1/a>1/b which implies that 5>2 then 1/5>1/2
3: Calculate total bowling costs
4:Given that n have the property that the sum of the digits of n and the sum of digits of n + 1 are odd numbers. FIND HOW MANY NUMBERS ARE IN N

source..
Content:


MATHEMATIC ASSIGNMENT 0005
INSTITUTION NAME:
STUDENT’S NAME:
DATE:
Question 1 Answer
Factoring method
x2-2x-15=0
Find the product and the sum of the equation
Product=1×-15=-15 and sum=-2
Expand the equation
x2-5x+3x-15=0
Factorise
x+3x-5=0
Set each equation equal to zero
x+3=0 or x-5=0
x=-3 or 5
Quadratic method
ax2+bx+c=0
x=-b±b2-4ac2a
x2-2x-15=0
Apply the quadratic formula
x=2±(-2)2-(4)(1)(-15)2(1)
x=2±4+602
x=2±642
Simplify the equation
x=2±82
x=2+82 or2-82=0
x=102 or-62=0
x=5 or-3
Completing Square method
Put the equation in the form of ax2+bx=-c
x2-2x=15
Make sure a=1
Using the value of b from new equation, addb22 to both sides of the equation to form a perfect square on the left side of the equation
c=-222=1
x2-2x+1=15+1
x2-2x+1=16
Find the square root on both sides
x-1=±4
Solve the result of the equation
x=4+1 or-4+1
x=5 or-3
Question 2 Answer
To prove this let assume that a=5 and b=2
Following the theorem
a>b then1a>1b which implies that 5>2 then15>12
Which negate the property. Thus a>b then1a>1b is false
Question 3 Answer
(a) Number of teenagers = 26 Computation of Cinema Costs Cinema charges = 4×26= £104 Transport charges to cinema = £90
Hence, total cinema cost = 104+ 90 = £194
Computation of bowling costs
Bowling alley charges = 5.5×26= £143 Transport charges to bowling = £45
Hence, total bowling costs = 143 + 45 = £188
Question 4 Answer
Given: n have the property that the sum of the digits of n and the sum of digits of n + 1 are odd numbers
To Find: How many numbers
Solution:
Case 1:  Sum of digit of n + sum of digits of n + 1   is odd
if sum of digit of n is even then sum of digit of n + 1 is even + 1
Hence, we get 2(even) + 1 = odd
if sum of digit of n is odd then sum of digit of n+1 is odd + 1
Hence, we get 2(odd) + 1 = odd
But there is case where n + 1 digit does not have sum 1 extra
whenever it changes like 9 to 10, 19 to 20 and so on
so, numbers which does not satisfy
9, 19, 29, 39 ,49, 59, 69, 79, 89   ( 99 is exceptional)
109, 119 , 129 , 139 , 149 , 159 , 169 , 179 , 189 , 199
up-to 1999  
Numbers f

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