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thanks!

14mn  =  55 - 7m - 2n

Add 2n to both sides:

14mn + 2n  =  55 - 7m

Factor:

n(14m + 2)  =  55 - 7m

Divide both sides by 14m + 2:

n  =  (55 - 7m) / (14m + 2)

If you replace m with 1:                                 (a guess; if it doesn't work, try 2 ... )

n  =  (55 - 7·1) / (14·1 + 2)  =  (55 - 7) / (14 + 2)  =  48 / 16  =  3

If m = 1, then n = 3

Angle:  x

Supplement to x:  180 - x

x = 4(180 - x) - 120

x = 720 - 4x - 120                         Now, add 4x to both sides:

5x  =  720 - 120

5x  =  600

x = 120

Angle = 120; supplement = 180 - 120 = 60.

Here to solve questions,

Here's a similar problem to think of:

One elephant can eat 2000 pounds of grass in 20 days.

He and his friend can eat it in 10 days.

How long does it take for the elephant's friend to eat alone?

Solution:

Again, elephant 1:1/20

elephant + friend: 1/10

Equation: 1/20+x=1/10

Take 1/20 to the other side, to get: 1/10-1/20=2/20-1/20=1/20 days

So, his friend can eat the grass alone in \(\boxed{20}\) days.

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Here's a different type of problem to think of.

A person can finish a poster completely in 20 minutes. His friend can finish the same poster in 10 minutes. How many minutes will it take if they work together:

1/20+1/10=1/20+2/20=3/20

1(whole)/3/20=20/3=\(6 \frac{2}{3}\)minutes.

Supplementary: 180

(180-x)*4-120

x=720-4x-120

5x=120

180-120=\(\boxed{60^\circ}\) degrees.

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

We can set up an equation, here:

First number= \(x\)

Second number=\(2x+3\)

So,  \(x+2x+3=66\)

Simplifying more, we have:

\(3x+3=66\)

\(3x=63\)

\(x=21\)

Since it's asking for the larger number, we simply do:

\(21*2+3=\boxed{45}\)

Thanks, Gavin, and tertre!

Thanks so much!

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.

The root of the problem is not one who gives to answers, but the ones who take it.

I agree with x^2 that his answers include brilliant solutions to helped users not only get the answer, but understand it. 

When supplying an answer, you should always include a solution. Whether or not they look at it, is completely up to them. Answering someone's math question is not cheating if you help them understand. It is them, who do not look at the solutions which is considered cheating. 

Hi Melody!
 

In that case, then this is possible. 

Using the same logic, a sphere is composed of mulitple cones. 

This is because a cone is considered a pyrimaid. 

Thank you for correcting my mistake,

Gavin

Hey Guest!

I have answered a few chemistry questions on this website.

It is supposed to be for math (I think).

If you want to ask a chemistry problem, ask away!

I will give it a go, no guarantees it will know how to do it. 

Also, FiestyGeco, a user posts multiple questions in 1 thread and just reads the user's answers, not solutions. 

I totally agree with you, Melody!

As I've been here for about 2 years now! (Wow!), I've seen homework questions been posted multiple times by users, but haven't seen the users type anything back after another user posts. They just read their answer, discarding their solution.

2. We think of a Pythagorean triplet, here:

a^2+b^2=c^2

So, 5^2+12^2=13^2

So, the answer is \(\boxed{2}\)

The current helps the frog swim downstream and hinders her swimming upstream. Let the frog's rate in still water be \(x\) and the current be \(y.\) Thus, the swimming downstream the frog's rate is \(x+y\) , while swimming upstream it is \(x-y\) . Now apply rate times equals distance: downstream, we have 

                    \((x+y)(2)=8\)

and upstream gives

                    \((x-y)(14)=8\)

Solving these equations, we find that \((x,y)=(16/7, 12/7).\)

Thus the frog's rate in still water is \(\boxed{\frac{16}{7}}\)miles per hour.

Hey ant101!

x = speed of frog

y = speed of current

We use (distance)=(rate)*(time) 

We can set up the system of equations:

\((x+y)\cdot2=8\\ (x-y)\cdot14=8\)

Solving for x, we get:

\(x=\frac{16}7\)

I hope this helped,

Gavin

knight=1/15

kinght+partner=1/10

partner=1/10-1/15=1/30

partner=30 days

If a knight can storm a castle in 15 days, he can storm 1/15 of a castle in one day. 

If he and his partner can do it in 10 days, they can storm 1/10 of a castle in one day. 

We can set up the equation: 

\(\frac{1}{15}+\frac{1}{x}=\frac{1}{10}\\x=30\)

It will take his partner 30 days to storm the castle alone. 

I hope this helped,

Gavin

Thank you very much. I envy you guys and i thnik how and when I will be intelligent like you guys... I started studying late. I am 22yrs. I feel like k*****g myself so I can reincarnate and correct my mistakes on time