Algebra Assignment: Solutions for Algebra Questions and Problems

This document presents a comprehensive solution set for an Algebra homework assignment. The solution covers a range of algebraic concepts, including solving equations, understanding variables and constants, calculating the steepness of a line, and working with divisors. The assignment addresses both linear and quadratic equations, providing detailed step-by-step solutions and explanations for each problem. The solutions include explanations of key concepts, the application of formulas, and the interpretation of results. The document provides answers to questions about variable and constant identification, quadratic formula application, and the divisibility of numbers. The assignment also includes an analysis of intersecting lines and curves. The document concludes with a discussion on the divisibility of concatenated numbers and finding remainders. This assignment is designed to help students master key concepts in Algebra.

Running head: ALGEBRA 1
Algebra
Name
Institution

ALGEBRA 2
Algebra
Question 1
If f ( t ) = a
t −a , then to get f ( a
t )we replace t with a
t as follows.
f ( a
t )= a
a
t −a
= a
a
t −a
× t
t = at
a−at = a ( t )
a(1−t)= t
1−t Question 2
To get the steepness of a line, we find the change in the y-coordinate as a ratio of the change in
the x-coordinate. That is, if we have two points on a line denoted by the coordinates
( t1 , y1 )∧(t2 , y2 ), the steepness a is defined by,
a= y2− y1
t2−t1
If we let y=b, then the line cuts the y-0axis at a point with the co-ordinates (0 , b). Then, from
the from figure 1 attached, if the two points on the line are (t , y )∧(0 , b) the equation of the line
can be derived as shown.
a= y2− y1
t2−t1
= y −b
x−0 = y−b
x
a= y−b
x
Making y the subject of the formula yields y=a t +b

ALGEBRA 3
Question 3
Part a
A variable is the parameter with an unknown value in an equation. That is, parameter x in the
equation a x2 +bx+ c=0. On the other hand, a constant is a value that does not change in an
equation. That is, its value is known. If we want to solve for the unknown variable in a quadratic
equation, we write the equation in the form a x2 +bx+ c=0 . Then, the corresponding values of a,
b, and c are plugged in the quadratic formula to solve for the unknown.
Part b
Yes, it works.
If we write y=ax2 in the form a x2 +bx+ c=0, we get a x2 +0 x+ 0=0 which means that a and c
are zero. Then we can insert the values in the quadratic formula to solve for the unknown.
Question 4

ALGEBRA 4
We can approach the question the same way. However, since we know that we can only get the
square root of non-negative integers, we can ignore the negative solution and have +2 as the only
solution to the equation.
Question 5
The information given is insufficient for us to deduce the age of the three children. Still, we
don’t know the street number of the house.
Question 6
Part a
The six-month old baby girl weighsw ( 6 )=1 5, pounds
Part b
An average 12-month old baby girl weighs 21 pounds.
Part c
w ( p )=1 4denotes that an average p-month old baby weighs 14 pounds.
Part d
An average 6-month old baby girl weighs 15 pounds. As a result, we expect the average 8-month
old baby girl to weigh more than 15 pounds. However, an 8-month old Schmeterling weighs
w ( 13 ). Therefore, she has a below average weight.
Question 7
Part a

ALGEBRA 5
300 has 18 divisors as follows:
1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 25, 30, 50, 60, 75, 100, 150, 300
360 has 24 divisors as follows:
1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 18, 20, 24, 30, 36, 40, 45, 60, 72, 90, 120, 180, 360
400 has 15 divisors as follows:
1, 2, 4, 5, 8, 10, 16, 20, 25, 40, 50, 80, 100, 200, 400
Part b
There is no positive smaller than 360 that has as many divisors as 360. Actually, 360 is divisible
by all integers between 1-10 except 7. As a result, it is divisible by multiples of the integers
between 1-10 except multiples of 7. Hence, it has many divisors than any other number below it.
Question 8
At the points of intersection between the line and the curve, y=x2. Then,
If x= A , y= A2 and if x=B , y=B2
The gradient, If m= ∆ y
∆ x = B2− A2
B− A =( B− A)(B+ A)
B−A =B+ A
But we know that the equation of a straight line is in the form y=mx+c
Therefore, the equation of the line becomes y=(B+ A )x +c
Question 9

ALGEBRA 6
It is true that such number is divisible by 13. For instance, if we concatenate 492 we get 492492.
If we divide 492492 by 13 we get 37884. Other proofs are shown in the table below.
3 digit number Concatenated form Divisible by 13?
199 199199 Yes
701 701701 Yes
572 572572 Yes
843 843843 Yes
667 667667 Yes
Therefore, it is true that any concatenated 3-digit number is divisible by 13.
Question 10
The smallest positive integer n=1618
I got the result using the EXCEL function “MOD” as shown in the attached screenshot.
The smallest positive integer n=1618. The second, third, and fourth columns represent the
remainder when n is divided by 5 , 8 ,∧49 respectively.