GYanggg

Yeah, I have realized. 

I was so caught up in writing the solution to the problem, 

I just made a careless error. 

Thanks for spotting it. 

A chord of a circle is a straight line segment whose endpoints both lie on the circle.

More generally, a chord is a line segment joining two points on any curve, for instance an ellipse.

A chord that passes through a circle's center point is the circle's diameter.

Every diameter is a chord, but not every chord is a diameter.

The first choice, OH is not a chord, it is the radius of the circle. 

SD is also not a chord, it is a tangent line to the circle. 

OR is too, not a chord, it is also a radius. 

RC is a chord, since a diameter is a chord. 

:) 

Hey Rollingblade!

We have two equations we can manipulate here:

First, let's change the second one:

\(x(x-1)(x-2)(x-3)\\ =[x(x-3)][(x-1)(x-2)]\\ =(x^2-3x)(x^2-3x+3)\)

Now, we can change the first one:

\(x=\frac{\sqrt5-3}{2}\\ 2x+3=\sqrt5\\ (2x+3)^2=5\\ 4x^2+12x+4=0\\ x^2+3x=-1\)

We now found the value of \(x^2+3x\), and can plug this into the original equation. 

\((x^2-3x)(x^2-3x+3)=-1(-1+3)=\boxed{-2}\)

I hope this helped,

Gavin. 

Hey tertre, 

The reason I did not fully solve was not because I was lazy. 

I wanted to make sure the guest knew what he was doing and understood the solution. 

This way, he will actually learn and not just copy the answer we give him. 

I was hoping that the guest was going to solve the problem I posed to him, not just receive the answer. 

For a parabola \(y=ax^2+bx+c\), the vertex's x is equal to \(\frac{-b}{2a}\)

The equation is given to us:

\(y=\frac34x^2 − 6x + 15\\ \)

The x is \(\frac{-(-6)}{2\cdot\frac34}\)

Then you can plug in for y, I hope this helped,

Gavin

Hey lightning!

If a quadratic equation has exxactly one real root, the discriminant is equal to zero. 

In the quadratic equation: \(ax^2+bx+c\), 

The discriminant is: \(b^2-4ac\).

In this specific problem, the discriminant is \(b^2-4c\), since \(a=1\).

From the information given, we can set up the systems:

\(b^2-4c=0,\\ b=c+1.\)

Solving the systems, we can substitute:

\((c+1)^2-4c=0\\ c=1\)

I hope this helped,

Gavin

Portia solved the quadratic equation \(x^2+(2\sqrt3)x+1=0\) by completing the square. In the process, she came up with the equivalent equation \((x+r)^2 = s\),where r and s are constants. What is s?

If you don't know how to complete a square, check out this website: https://www.mathsisfun.com/algebra/completing-square.html

However, if you're just stuck on this question, you can view my solution here:

We can complete the square by adding 2 to both sides. 

\(x^2+(2\sqrt3)x+3=2\\ (x+\sqrt3)^2=2\\ \)

From this, we can tell:

\(r=\sqrt3\\ s=2\)

The answer you seek is 2, 

I hope this helped,

Gavin

Hey GAMEMASTERX40!

The formula to find the area of a square is L^2, where L is the side length. 

Therefore, to find the side length of square #1, we want to find the square root of x^2+4x+4.

We can factor this, since it’s a perfect square:

x^2+4x+4=(x+2)^2, the side of the square is x+2. 

Therefore, the side length of square #3 is twice of this, or 2x+4. 

The area of square #3 must be (2x+4)^2.

I hope this helped,

Gavin

Hey GAMEMASTERX40!

We want to find the first four terms of the recursive sequnce defined above. 

The first term is already given, since it says a₁ = -3, so we know the first term is negative three. 

The second terms is defined by the second rule. It states that the next term is 6 less then the term before. 

So if the first term is -3, the second is -3-6 = -9

We continue this rule:

-3, -9, -15, -21

I hope this helped,

Gavin

Hey superman!

The goal is to count in base 3 using only binary digits.

The \(100^{\text{th}}\) smallest positive binary integer is \(100 = 1100100_2,\) 

so the \(100^{\text{th}}\) smallest positive integer that can be written with only the binary digits is

 \(1100100_3 = \boxed{981}\).

I hope this helped,

Gavin.