Determine the equation of the line that passes through 

Let the even number be 2n. Its consecutive even number will be 2n+2
Finding the product of 2n and (2n+2)

2n(2n+2) = 2n(2n) + 2n(2)

2n(2n+2) = 4n² + 4n
The given conjecture is “The product of any two consecutive even numbers is 1 less than a perfect square”

So, add 1 to 4n² + 4n

4n² + 4n +1 = (2n)² + 2(2n)(1) + 1²

= (2n + 1)²
So, we can see that the perfect square of 2n+1 is 4n² + 4n +1 and 1 less than 4n² + 4n +1 is a product of any two consecutive even numbers.
Final Answer:
Hence, the given conjecture is right and we have proved it by Deductive reasoning.

Explanation:
We have to identify the linear polynomial from the given four options in the question
Step 1 of 1:
A linear polynomial is defined as any polynomial expressed in the form of an equation of .
From the option We can see that option B is in the form of 
So, Option B is correct.

Answer:

• Hint:
• Mid-point theorem:
• According to mid-point theorem, in triangle, the line segment which joins the midpoint of two sides is parallel to third side and is equal to half of third side,

• Step by step explanation: 
• Given:

From figure,
BM = MC,
hence M is midpoint of BC.
BN = NA,
hence N is midpoint of AB.
MN = x
AC = 24

• Step 1:

In 
The line segment MN joins midpoint of AB and BC
So,
According to midpoint theorem,
MN is parallel to AC and
MN = 
MN = 
MN = 12
∴ x = 12.
Final Answer:
x = 12.

Explanation:

• We have to identify the cubic polynomial from the given four options.

Step 1 of 1:
We know that a cubic polynomial is a polynomial with degree 3.
Option A:
x + 3
The degree of this polynomial 1.
Option B:

The degree of the polynomial is 4.
Option C:

The degree of this polynomial is 3.
Option D:

The degree of this polynomial 2.

(mx + ny)² can be written as (mx + ny)(mx + ny)
(mx + ny)(mx + ny) = mx(mx + ny) + ny(mx + ny)
=  mx(mx) +  mx(ny) + ny(mx) + ny(ny)
= m²x² + mnxy + mnxy + n²y²
= m²x² + 2mnxy + n²y²
Now, m²x² + 2mnxy + n²y² = 4𝑥² + 12𝑥𝑦 + 9𝑦²
Comparing both sides, we get
m² = 4, n = 9, 2mn = 12
So, m = +2 or -2 , n = +3 or -3
Considering 2mn = 12, there are two combinations possible

1. m = +2 and n = +2
2. m = -2 and n = -2

Final Answer:
Hence, the values of (m, n) are (2,2) and (-2,-2).

Answer:

• Step by step explanation:

○    Given:
correct answer = 12 points
incorrect answers = -5 points
School A starts with 165 points and gives the same number of correct and incorrect answers.
School B starts with 65 points and gives no incorrect answers and the same number of                           correct answers as school A
○    Step 1:
○    Let the number of correct answers given by school A be x.
So, the number of incorrect answers is also x.
At school A starts with 165 points. After giving x correct and a incorrect answers points will be
165 + 12x - 5x

School B starts with 65 points. Schools are given the same number of correct answers as                       school A and no incorrect answers. So, their points will be

                65 + 12x
○    Step 2:
○    As both schools tied
∴ 165 + 12x - 5x = 65 + 12x

• Final Answer:

Correct option:
Option B. 165 + 12x - 5x = 65 + 12x

Answer:

• Hints:
• Perpendicular bisector theorem
• According to perpendicular bisector theorem, in the triangle, any point on perpendicular bisector is at equal distance from both end points of the line segment on which it is drawn.

• Step by step explanation: 
• Given:

AB = 2x.
AC = 4x – 4
AD is perpendicular bisector at BC.

• Step 1:
• In  

AD is perpendicular bisector.
A is point on AD
A is equidistant from B and C.
So,
AB = AC
2x = 4x – 4
4 = 4x – 2x
4 = 2x

x = 2

• Final Answer: 

             Hence, x = 2.
Perpendicular bisector theorem is used.

Answer:

• Hints:

• Distance between two points having coordinates (x₁, y₁) and (x₂, y₂) is given by formula:

• Distance =

• Step by step explanation: 
  • Step 1:
  • Let triangle ABO,

where,
O = (0, 0)
A = (2a, 0)
B = (0, 2b).

• Step 1:
• Let triangle ABO, where:

The midpoint of BC is given by,


So, the perpendicular bisector will intersect BC at M (a, b).
Equation of line BC is




And point M (a, b) satisfy above equation


b = b
Hence, point M (a, b) lies on BC.

• Final Answer: 

Hence, circumcentre of right angle triangle lie on midpoint of hypotenuse.

Answer:

• Hint:

○    The concept used in the question is of the quadratic equation.

• Step by step explanation:

○    Given:
Health club 1: Membership R s 22 and Monthly fee R s 24.50
Health club 2: Membership R s 47.00 , monthly fee R s 18.25.
○    Step 1:
○    Let the number of months after which the total cost is equal be x.
So, for Health club 1:
After x months, total cost will be =
Membership Rs 22 + fee for x months
 22 + 24.50x
Health club 2:
After x months, total cost will be =
Membership R s 47 + fee for x months
 47 + 18.25x
○    Step 2:
○    Equalize both costs to get number of months
22 + 24.50x = 47 + 18.25x
24.50x - 18.25x= 47 - 22
6.25x = 25
x = 
x = 4

• Final Answer:

Hence, after 4 months the total cost of each health club will be equal.

Hint:
Gradient is also called the slope of the line. The slope intercept form of the equation of the line is y = mx + c, where m is the slope of the line and c is the y-intercept. First we convert the given equation in this form. Further, compare the equation with the standard form to get the slope and the y-intercept.
Step by step solution:
The given equation of the line is
x + 2y = 14
We need to convert this equation in the slope-intercept form of the line, which is
y = mx + c
Rewriting the given equation, we have
2y = 14 - x
Dividing by 2, we get

Simplifying, we get

Comparing the above equation with , we get

Thus, we get
Gradient = 
y-intercept = 7
Note:
We can find the slope and y-intercept directly from the general form of the equation too; slope =   and y-intercept = , where the general form of equation of a line is ax + by + c = 0. Using this method, be careful to check that the equation is in general form before applying the formula.

Explanation:

• We have given a constant
• We have to find how we can express any constant 𝑘 in the variable form without changing its value.

Step 1 of 1:
We have given a constant k
We have to find how we can express any constant 𝑘 in the variable form without changing its value
We know that x⁰ = 1
So After multiplying it with any constant, it will not change its value.
So,kx⁰ is the answer
Hence, Option C is correct.

Answer:

• To prove: 
• Perpendicular Bisector Theorem:

• Proof: 
• Statement:
• According to perpendicular bisector theorem, in the triangle, any point on perpendicular bisector is at equal distance from both end points of the line segment on which it is drawn.

• Step 1:

Let consider given figure,

Let arbitrary point C on perpendicular bisector.
In which,
CD is perpendicular bisector on AB.
Hence,
AD = DB
CD = CD (common)

So, according to SAS rule

Hence,
CA = CB
So, any point on perpendicular bisector is at equal distance from end points of line segment.
Hence proved.

Explanation:

• We have to define monomial by giving an example.

Step 1 of 1:
A monomial is an expression with only one term.
The examples are: 3x, 6y

Hint:
We are given two points and we need to find the equation of the line passing through them. The equation of a line passing through two points (a, b) and (c, d) is

Step by step solution:
Let the given points be denoted by
(a, b) = (-3, 1)
(c, d) = (2, -14)
The equation of a line passing through two points (a, b) and (c, d) is

Using the above points, we have

Simplifying the above equation, we have


Cross multiplying, we get
5(y + 14) = -15(x - 2)
Expanding the factors, we have
5y + 70 = -15x + 30
Taking all the terms in the left hand side, we have
15x + 5y + 70 - 30 = 0
Finally, the equation of the line is
15x + 5y + 40 = 0
Dividing the equation throughout by 5, we get
3x + y + 8 = 0
This is the required equation.
Note:
We can simplify the equation in any other way and we would still reach the same equation. The general form of an equation in two variables is given by ax + by + c = 0,, where a, b, c are real numbers. The student is advised to remember all the different forms of a line, like, slope-intercept form, axis-intercept form, etc.