GYanggg

Once again, we use the 30 - 60 -90 rule. 

The hypotenuse is twice the length of the side opposite of the 30 degree angle.

Therefore the side opposite of the 30 degree angle is 12/2 = 6 

We can use the Pythagorean theorem, to find the length of a. 

12^2 - 6^2 = a^2

a^2 = 108

a = √108 = 6√3

I hope this helped,

Gavin

In a 30 - 60 - 90, triangle, the side opposite of the 30 degree angle is half of the hypotenuse.

We don’t have to treat this problem as 3D geometry, we can just pay attention to the side facing us. Now it is just a 2D geometry problem. 

Using the rule I stated at the top of this post, the hypotenuse is twice the side opposite of the 30 degree angle, which is 3 meters long. 

Therefore, the hypotenuse is 3*2 = 6 

Since we know 2 sides of a right triangle, we can find the third. 

using the Pythagorean theorem, x^2 = 6^2 - 3^2 = 30 

x = √30

I hope this helped,

Gavin

To write this out algebraically, we can assign variables to the two numbers. 

First number = x, Second number = y

The product of the two numbers is x*y or just xy

The sum of the two numbers is (x+y) 

The sum of the product and the sum of two positive integers can be expressed as: 

xy + x + y, and this is equal to 454, xy + x + y = 454 

We can solve for all values of x and y, we get: 

xy + x + y = 454

We add 1 to both sides so we can factor: 

xy + x + y + 1 = 455

Then we can factor:

x(y+1)+y+1 = 455

(y+1)(x+1) = 455 = 5 * 7 * 13

Their product is the largest only when x = 90 and y = 4 

(90*4)(90+4)=33840

Other pairs is not as big, for example, when x = 12 and y = 34

(34*12)(34+12)=18768

To answer your question, the largest possible value is 33840, 

I hope this helped,

Gavin

Good to see you back Geno!

No problem, glad I can help!

No problem, glad I can help!

Hi Guest!

We use: \(x^3-y^3=(x-y)(x^2+xy+y^2) \)

We already know \( (x-y)\) and \(x^2+y^2\)

We can plug the known values in, and we get:

\(6(24+xy)=x^3-y^3\)

To solve for xy, we use: \((x-y)^2=x^2-2xy+y^2\)

This means that \(6^2=24-2xy\)

Solving for xy, we get:

\(36=24-2xy\\ 12=-2xy\\ xy=-6\)

Now we can plug this into the original expression:
\(6(24-6)=x^3-y^3\\ x^3-y^3=108\)

I hope this helped,

Gavin

Hi Guest!

We start by isolating the y term:

6y + 2x - 2x = 12 - 2x

6y = 12 - 2x

Now we divide both sides by 6:

6y ÷ 6 = (12 - 2x) ÷ 6

y = 2 - x/3

Writing it in the slope intercept form: 

y = -1/3x + 2

I hope this helped,

Gavin

Given that \(a(a+2b) = \frac{104}3, b(b+2c) = \frac{7}{9}, \text{and}\ c(c+2a) = -7, \text{find}\ |a+b+c|.\)

We distribute the variables:

\(a^2+2ab=\frac{104}3,b^2+2bc=\frac79, \text{and} \ c^2+2ac=-7\)

Add the three equations:

\(a^2+b^2+c^2+2ab+2bc+2ac=\frac{104}{3}+\frac79-7\)

We recognize the left side of the equation as the expanded form of \((a+b+c)^2\)

We can rewrite the equation in the form like this:

\((a+b+c)^2=\frac{256}9 \)

Since the problem asks for \( |a+b+c|\), we take the square root:

\(\sqrt{(a+b+c)^2}=\sqrt{\frac{256}9}\\ a+b+c=\pm\frac{16}3 \)

Therefore, \( |a+b+c|=\boxed{\frac{16}3}\) 

I hope this helped,

Gavin

Hi Rollingblade!

Melody made a similar answer here: https://web2.0calc.com/questions/weekly-cycle-challenge-2-help 

I was thinking about Melody's answer and was thinking about how to generalize it. Here is what I have:

Using the hexagon problem: 

The area of the hexagon is equal to the sum of the six triangles constructed from connecting the point inside the hexagon to the 6 vertices. 

\(S_{abcdef}=\frac12m(a+b+c+d+e+f)\)

The area can also be expressed as the sum of the areas of the 6 equilateral triangles. 

\(S_{abcdef}=\frac12\cdot6\cdot{m}\cdot{h}\)

Since these two expressions are equal, we get:

\(6h=a+b+c+d+e+f\)

This is true for any polygon. 

n-sided polygon:

\(n\cdot{h}=\overbrace{a+b+c+d+\dots+x+y+z}^\text{n-sides}\)

Where h is the height, and a - z are the known perpendicular lines from any point.