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ah i see! i was making it way too difficult :] thanks for the explanation !

Since you know the number of apples, all you have to do is divide it by its ratio of 5: 35 / 5 =7 . Then you would simply multiply this 7 by the ratio of pears and that will give you the number of pears in the box.

So, 7 x 3 =21 pears. This is because: 35 apples:21 pears = The ratio of: 5:3

A regular dodecahedron \(P_1 P_2 P_3 \dotsb P_{12}\) is inscribed in a circle with radius 1.
Compute  \((P_1 P_2)^2 + (P_1 P_3)^2 + \dots + (P_{11} P_{12})^2\).
(The sum includes all terms of the form \((P_i P_j)^2\), where \(1 \le i < j \le 12\).

The \((P_1 P_2)\) isn't multiplication, it shows the distance from point one to two.

Let g(x) = f(x)¹⁶.     Write g(x) as  g(x) = g(x)_eventerms + g(x)_oddterms.

g(1) = g(1)_eventerms + g(1)_oddterms  ...(1)

g(-1) = g(-1)_eventerms + g(-1)_oddterms  

g(-1) =  g(1)_eventerms - g(1)_oddterms  ...(2)

(1) + (2):  g(1) + g(-1) = 2*g(1)_eventerms 

so: g(1)_eventerms = (g(1) + g(-1))/2

or g(1)_eventerms = (f(1)¹⁶ + f(-1)¹⁶)/2

Thanks Heureka,

DragonLord,

I am wondering why you are not posting your questions directly onto the forum?

You could post on the forum and then send a link to Heureka asking if he could help you.

That would be the usual way to do it if you want a specific person to answer.

A square is inscribed in a right triangle, as shown below.

The legs of the triangle are 2 and 3. 

Find the side length of the square.

\(\text{Let $A=(0,\ 0)$ } \\ \text{Let $B=(x_b,\ y_b)=(\sqrt{13},\ 0)$ } \\ \text{Let $C=(x_c,\ y_c)=\Big(2\cos(A),\ 2\sin(A)\Big)$ } \\ \text{Let $ \cos(A)=\dfrac{2}{\sqrt{13}},\ \sin(A)=\dfrac{3}{\sqrt{13}}$ } \)

\(\begin{array}{|lrcll|} \hline & \tan{\varphi} &=& \dfrac{2}{2+3} \\\\ & \tan{\varphi} &=& \dfrac{2}{5} \\\\ \text{line }1: & y &=& \tan{\varphi}*x \\\\ & \mathbf{y} &=& \mathbf{\dfrac{2}{5}*x} \\ \\ \hline \text{line }2: & \dfrac{y-y_b}{x-x_b} &=& \dfrac{y_c-y_b}{x_c-x_b} \\\\ &&& \boxed{x_b=\sqrt{13},\ y_b = 0,\quad x_c=\dfrac{4}{\sqrt{13}},\ y_c = \dfrac{6}{\sqrt{13}} } \\\\ & \dfrac{y}{x-\sqrt{13}} &=& \dfrac{\dfrac{6}{\sqrt{13}}}{\dfrac{4}{\sqrt{13}}-\sqrt{13}} \\\\ & \dfrac{y}{x-\sqrt{13}} &=& \dfrac{6}{4-13} \\\\ & \dfrac{y}{x-\sqrt{13}} &=& -\dfrac{2}{3} \\\\ & \mathbf{y} &=& \mathbf{-\dfrac{2}{3}(x-\sqrt{13})} \\ \hline \end{array}\)

The intersection point of both lines:

\(\begin{array}{|rcll|} \hline y= \dfrac{2}{5}*x &=& -\dfrac{2}{3}(x-\sqrt{13}) \\\\ \dfrac{2}{5}*x &=& -\dfrac{2}{3}(x-\sqrt{13}) \\\\ \dfrac{9}{10}*x &=& -(x-\sqrt{13}) \\\\ \dfrac{9}{10}*x &=& -x+\sqrt{13} \\\\ x+ \dfrac{9}{10}*x &=& \sqrt{13} \\\\ \dfrac{19}{10}*x &=& \sqrt{13} \\\\ x &=& \dfrac{10\sqrt{13}}{19} \\\\ y = s &=& \dfrac{3}{5}x \\\\ s &=& \dfrac{3}{5}*\dfrac{10\sqrt{13}}{19} \\\\ s &=& \dfrac{6}{19} \sqrt{13} \\ \mathbf{s} &=& \mathbf{1.13859513962} \\ \hline \end{array}\)

The side length of the square is \(\mathbf{\dfrac{6}{19} \sqrt{13} = 1.13859513962}\)

CPhill works 24 hours a day and solve 60 problems per hour, when will he get bored?

How many hours passed?

For visulation:

7 - 8 am, 8 - 9 am, 9 - 10 am, 10 - 11 am, 11 - 12 am, 12 - 1 pm, 1 - 2 pm.

7 hours

1. x + 7 * 2 = 10

2. x + 14 = 10

3. x = -4

-4 Celcius

Hello Guest! Make an account! 

Hypotenuse of big triangle: \(\sqrt{13}\)

This is just a compilation of a bunch of equations!

By Pythagorean Theorem:

(1) \((2-b)^2+(3-d)^2=b^2-a^2\)

(2) \(b^2-a^2=d^2-c^2\)

(3) \(d^2-c^2=\sqrt{13}-a-c\)

By similarity where \(s\) is the side length of the square:

(4)  \(\frac{a}{b}=\frac{d}{s}\)

(5) \(\frac{2}{3}=\frac{2-b}{3-d}\)

(6) \(\frac{2}{3}=\frac{s}{c}\)

(7) \(\frac{2}{3}=\frac{a}{s}\)

General:

(8) \(a+c+s=\sqrt{13}\)

y=3

then y=9

What was the change? Oh sure, it was multiplied by 3.

That means x must be multiplied by 3 too!

3 * 10 = 30

yay

Imagine two right triangles. They are similar because the sun shines at the same angle or the problem wouldn't make sense with the info given.

1. \(\frac{\text{shadow}}{\text{height}}=\frac{\text{shadow}}{\text{height}}\)

2. \(\frac{1.8}{3}=\frac{12}{x}\)

3. Solve for \(x\)

Super-hard-impossible bonus question:

Given the information in the problem.

If you were able to stand on top of the mail box and look STRAIGHT up into the sky, the sky can represent a 2D coordinate plane. Where would the Sun be located in (x,y)?

Consider everything you know (sorry physicists)!

Hello Guest!  I cannot correctly interpret the problem based on the format. I CAN give you a method to solve for one side length of the dodecagon, which no doubt can help you solve the problem yourself.

Interpret the problem

It's a pizza!
Look at ONE of the slices of the pizza. It has a vertex angle of 30 degress. (We calculate this by doing 180 - (360/12) = 150, then 180-150)

We know that the two long sides of the pizza is 1.

By law of cosines:

c² = a² + b² - 2ab*cos(C)

Plug in and solve:

c² = 1 + 1 - 2cos(30)

c² = 2 - 2cos(30)

c² = 1.691497100224

The problem cannot make sense as it ruins the pattern though.

So does it mean it includes EVERY diagonal and EVERY side length? Try to look at the original problem on the screen and match it with mine.

Set Up:

Length = L

Width = W

L = W + 12

Area = LW

Simplify:

W(W+12) = 85

W² + 12W - 85 = 0

Factor:

(W + 17)(W - 5) = 0 

Solve:

W = -17 or W = 5

I will leave the rest up to you. (Hint: lengths can't be negative.)

Assuming the problem means that the basketball has a radius of 4 inches when it is FILLED up, then here we go:

Sphere: \(\frac{4}{3}\pi{r^3}\)

Substitute and solve for volume:

\(\frac{256}{3}\pi\text{ cubic inches}\)

We divide:

\(\frac{256/6}{3}\pi\)

We evaluate:

44.6804288510548372040483

We round:

44.7 seconds

i solved it already, but thanks for trying to help! the answer was 120.

Hmmm, I couldn't find a geometric way to solve this within the 1 minute I looked at this problem. I am very inpatient!

So let us use coordinates!

____________________________________________________________________________________________

Find ST

ST = 40/3 by simple algebra I think you can do at your skill level judging by the difficult of this problem.

Let QR = \(a\)

Slope of PT: \(-\frac{\frac{40}{3}}{a}\)

Y-intercept of PT: 20

Equation: \(y=-\frac{\frac{40}{3}}{a}x+20\)

Slope of QS: (20/a)

Y intercept of QS: 0

Equation: \(y=\frac{20}{a}x\)

Substitute

\(-\frac{\frac{40}{3}}{a}x+20=\frac{20}{a}x\)

\(20 =\frac{\frac{100}{3}}{a}x\)

\(20a=\frac{100x}{3}\)

\(60a=100x\)

\(3a=5x\)

\(\frac{5}{3}x=a\)

Interpret this:

\(x\) is the x-coordinate of the solution of we solved for the location of the intercept at point U. That means \(x\) is the length of QV. We know \(a\) is the length of QR.

That means QV is three-fifths the length of QR.

We know that QUV is similar to QSR by AA similarity through a series of proofs.

Since we know the proportion of the sides, we can solve for UV:

Solve:

20 * (3/5) = 12

Ta-da! Mathz! 

I also don't think it is a series. It's supposed to be lengths from point to point within the dodecagon.

Hello guest!

Here is a hint: Can you prove triangle QPT and RQS similar? If so, you can use proportions to solve!

I just can't find a way to use proportions to solve it!

I think it is supposed to include diagonals, yes it looks like that.

Is this what it looks like?

\(\text{A regular dodecagon }P_1, P_2, P_3, ... P_12 \text{is inscribed in a circle with radius 1. }\)

\(\text{Compute } (P_1 P_2)^2 + (P_1 P_3)^2 + ... + (P_{11} P_{12})^2. \text{(The sum includes all terms of the form } (P_i P_j)^2, \text{where 1≤ i < j ≤ 12)}\)

In the series, are you sure the second term is \((P_1P_3)^2\)? It ruins the pattern!

Sorry for the confusion, and thank you for taking the time to help me!

Yes, it is the length

thx man and I love my name to XD