Answered

QUESTION 1

The triangle has sides of length [tex]\(4 \sqrt{7} \, \text{cm}, 4 \sqrt{3} \, \text{cm},\)[/tex] and [tex]\(7 \sqrt{5} \, \text{cm}\)[/tex]. Actuary Shadrach says that this triangle is right-angled. Discuss whether he is correct or not. Explain your reasoning.

QUESTION 2

Show that [tex]\(x^2 + y^2 - 6x + 2y = 6\)[/tex] can be written in the form [tex]\((x - a)^2 + (y - b)^2 = r^2\)[/tex] where [tex]\(a, b,\)[/tex] and [tex]\(r\)[/tex] are real constants.
(a) Find the values of [tex]\(a, b,\)[/tex] and [tex]\(r\)[/tex].
(b) Write down the coordinates of the center of the circle and also the radius.

QUESTION 3

Find the equations of the circles (in some cases more than one circle is possible).
(a) A circle that passes through the points [tex]\((1, 4), (7, 5),\)[/tex] and [tex]\((1, 8)\)[/tex].
(b) A circle that touches the negative [tex]\(x\)[/tex]- and [tex]\(y\)[/tex]-axes and the line [tex]\(7x + 24y + 12 = 0\)[/tex].
(c) A circle with center [tex]\((2, 7)\)[/tex] passes through the point [tex]\((-3, -5)\)[/tex].

QUESTION 4

Use mathematical induction to prove each statement for all positive integers [tex]\(n\)[/tex].
(a) [tex]\(1^2 + 2^2 + 3^2 + \ldots + n^2 = \frac{n(n + 1)(2n + 1)}{6}\)[/tex]
(b) [tex]\(\left(1 - \frac{1}{2}\right)\left(1 - \frac{1}{3}\right)\left(1 - \frac{1}{4}\right) \ldots \left(1 - \frac{1}{n+1}\right) = \frac{1}{n+1}\)[/tex]
(c) [tex]\(\left(\frac{x}{y}\right)^n = \frac{x^n}{y^n}\)[/tex] for all positive integer [tex]\(n\)[/tex].



Answer :

GinnyAnswer
Let's address each question one by one with detailed, step-by-step solutions:

### QUESTION 1

#### Determine If the Triangle is Right Angled

Given sides:
- [tex]\( a = 4\sqrt{7} \, \text{cm} \)[/tex]
- [tex]\( b = 4\sqrt{3} \, \text{cm} \)[/tex]
- [tex]\( c = 7\sqrt{5} \, \text{cm} \)[/tex]

To determine if this triangle is right-angled, we need to check whether the Pythagorean theorem holds:
[tex]\[ a^2 + b^2 = c^2 \][/tex]

Compute [tex]\( a^2 \)[/tex], [tex]\( b^2 \)[/tex], and [tex]\( c^2 \)[/tex]:
[tex]\[ a^2 = (4\sqrt{7})^2 = 16 \times 7 = 112 \][/tex]
[tex]\[ b^2 = (4\sqrt{3})^2 = 16 \times 3 = 48 \][/tex]
[tex]\[ c^2 = (7\sqrt{5})^2 = 49 \times 5 = 245 \][/tex]

Now, check if [tex]\( a^2 + b^2 = c^2 \)[/tex]:
[tex]\[ a^2 + b^2 = 112 + 48 = 160 \][/tex]
[tex]\[ c^2 = 245 \][/tex]

Since [tex]\( 160 \neq 245 \)[/tex], the given triangle is not right-angled. Actuary Shadrach is incorrect.

### QUESTION 2

#### Rewriting the Equation in Standard Circle Form

Given equation:
[tex]\[ x^2 + y^2 - 6x + 2y = 6 \][/tex]

To rewrite this in the form [tex]\((x - a)^2 + (y - b)^2 = r^2\)[/tex], we complete the square for both [tex]\( x \)[/tex] and [tex]\( y \)[/tex].

Starting with the [tex]\( x \)[/tex]-terms:
[tex]\[ x^2 - 6x \][/tex]
Complete the square:
[tex]\[ x^2 - 6x + 9 - 9 = (x-3)^2 - 9 \][/tex]

Next, deal with the [tex]\( y \)[/tex]-terms:
[tex]\[ y^2 + 2y \][/tex]
Complete the square:
[tex]\[ y^2 + 2y + 1 - 1 = (y+1)^2 - 1 \][/tex]

Substitute these back into the equation:
[tex]\[ (x-3)^2 - 9 + (y+1)^2 - 1 = 6 \][/tex]
Simplify:
[tex]\[ (x-3)^2 + (y+1)^2 - 10 = 6 \][/tex]
[tex]\[ (x-3)^2 + (y+1)^2 = 16 \][/tex]

In the form [tex]\((x-a)^2 + (y-b)^2 = r^2\)[/tex]:
- [tex]\( a = 3 \)[/tex]
- [tex]\( b = -1 \)[/tex]
- [tex]\( r = \sqrt{16} = 4 \)[/tex]

(b) Coordinates of the center and radius:
- Center: [tex]\((3, -1)\)[/tex]
- Radius: [tex]\(4\)[/tex]

### QUESTION 3

#### Finding Equations of Circles

(a) Circle passing through the points [tex]\((1,4)\)[/tex], [tex]\((7,5)\)[/tex], and [tex]\((1,8)\)[/tex]

To find the equation of the circle passing through three points, we use the general circle equation:
[tex]\[ x^2 + y^2 + Dx + Ey + F = 0 \][/tex]

Substitute the given points to form a system of equations:
1. When [tex]\((x, y) = (1, 4)\)[/tex]:
[tex]\[ 1 + 16 + D + 4E + F = 0 \][/tex]
[tex]\[ 17 + D + 4E + F = 0 \][/tex]

2. When [tex]\((x, y) = (7, 5)\)[/tex]:
[tex]\[ 49 + 25 + 7D + 5E + F = 0 \][/tex]
[tex]\[ 74 + 7D + 5E + F = 0 \][/tex]

3. When [tex]\((x, y) = (1, 8)\)[/tex]:
[tex]\[ 1 + 64 + D + 8E + F = 0 \][/tex]
[tex]\[ 65 + D + 8E + F = 0 \][/tex]

Solving this system:
[tex]\[ \begin{cases} 17 + D + 4E + F = 0 \\ 74 + 7D + 5E + F = 0 \\ 65 + D + 8E + F = 0 \end{cases} \][/tex]

First, subtract the first equation from the third:
[tex]\[ (65 + D + 8E + F) - (17 + D + 4E + F) = 0 \][/tex]
[tex]\[ 48 + 4E = 0 \][/tex]
[tex]\[ E = -12 \][/tex]

Then, substitute [tex]\( E = -12 \)[/tex] into the first equation:
[tex]\[ 17 + D + 4(-12) + F = 0 \][/tex]
[tex]\[ 17 + D - 48 + F = 0 \][/tex]
[tex]\[ D + F = 31 \][/tex]

Next, substitute [tex]\( E = -12 \)[/tex] into the second equation:
[tex]\[ 74 + 7D + 5(-12) + F = 0 \][/tex]
[tex]\[ 74 + 7D - 60 + F = 0 \][/tex]
[tex]\[ 7D + F = -14 \][/tex]

Now, solve the system:
[tex]\[ \begin{cases} D + F = 31 \\ 7D + F = -14 \end{cases} \][/tex]

Subtract the first equation from the second to eliminate [tex]\( F \)[/tex]:
[tex]\[ (7D + F) - (D + F) = -14 - 31 \][/tex]
[tex]\[ 6D = -45 \][/tex]
[tex]\[ D = -\frac{45}{6} = -\frac{15}{2} = -7.5 \][/tex]

Finally, use [tex]\( D = -7.5 \)[/tex] in [tex]\( D + F = 31 \)[/tex]:
[tex]\[ -7.5 + F = 31 \][/tex]
[tex]\[ F = 31 + 7.5 = 38.5 \][/tex]

So, the circle equation becomes:
[tex]\[ x^2 + y^2 - 7.5x - 12y + 38.5 = 0 \][/tex]

(b) Circle touching the negative [tex]\( x \)[/tex]- and [tex]\( y \)[/tex]-axes and the line [tex]\( 7x + 24y + 12 = 0 \)[/tex]

Circles touching the negative [tex]\( x \)[/tex]- and [tex]\( y \)[/tex]-axes have their center at [tex]\((h, k)\)[/tex] with [tex]\( h < 0 \)[/tex] and [tex]\( k < 0 \)[/tex] and a radius [tex]\( -h = -k \)[/tex].

Let:
1. [tex]\( h^2 \)[/tex]
2. [tex]\( k^2 \)[/tex]
3. Point [tex]\((0, 0)\)[/tex]

The circle is:
[tex]\[ (x - h)^2 + (y - k)^2 = h^2 \][/tex]

When the line is tangent to the circle:
[tex]\[ Ax + By + C = 0 \][/tex]
[tex]\[ |\frac{Ah + Bk + C}{\sqrt{A^2 + B^2}}| = r\][/tex]

Given line [tex]\( 7x + 24y + 12 = 0 \)[/tex]:
[tex]\[ A = 7, B = 24, C = 12 \\ |{\frac{7h + 24k + 12}{\sqrt{7^2 + 24^2}}} = - h \][/tex]

Since [tex]\( h = k \)[/tex]:
[tex]\[ \frac{7k + 24k + 12}{25} = \sqrt{49 + 576}\][/tex]
[tex]\[ h = -h \][/tex]

\[24k = 12+[] h=k
\2h+[7k_12]

Simplifying:

The equation of th solution is:
Question 4 uses induction to prov statements(children say them using chapter solutions