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First thing we know BC = BK + CK = 14.

Then we use Pythagorean theorem twice to get:

\(AB = \sqrt{AK^2 + BK^2} = 10\)

\(AC = \sqrt{AK^2+CK^2} = 6\sqrt2\)

Therefore,

Perimeter = AB + AC + BC = 14 + 10 + 6√2 = 24 + 6 √ 2

\(\sin A = \dfrac{a}{c}\\ \cos A = \dfrac{b}{c}\\ \tan A = \dfrac{a}{b}\\\)

:)

Did you miss 8 Heureka ?

It seems fairly clear from the following diagram that it is not always true.  The specified angle bisectors clearly won't intersect on AD. 

The proposition may well be true for a non-intersecting diagram.

.

1.4E+07 usually means 1.4*10⁷ = 14 000 000, or 14 million.  

But, isn't 42 + 13  mod 34 = 21 and 76 +13 mod 34 = 21 ????

\(\text{looking at the 2nd half of the show first}\\ \text{the songs will be AAAA!A}\\ \text{there are }4!=24 \text{ ways to arrange the 4 acoustic songs, and}\\ \text{10 ways to select the !A song at the end}\\ \text{this is then 240 ways to arrange the 2nd half of the show}\\ \text{the first half of the show is then 5 non-acoustic songs chosen from 9}\\ \text{since we care about order there are }\dbinom{9}{5}5! = 15120\\ \text{ arrangements for the first half} \\ \text{so in total we have}\\ 240 \times 15120 = 3628800 \text{ possible concert arrangements}\)

A regular octagon withside length 4 cm is concentricwith a circle inside,

If the area of the circle is equal to the area to the shaded region between the shapes,

what is the radius of the circle?

\(\text{Let $s =$ octagon side length $ = 4~ cm$}\\ \text{Let $a =$ octagon apothem length}\)

\(\begin{array}{|rcll|} \hline A_{\bigcirc} &=& A_{\text{octagon}} - A_{\bigcirc} \\ 2A_{\bigcirc} &=& A_{\text{octagon}} \\ \mathbf{A_{\bigcirc}} &=& \mathbf{\dfrac{A_{\text{octagon}}} {2}} \\ \hline \end{array}\)

apothem

\(\begin{array}{|rcll|} \hline a &=& \dfrac{s}{2}\cdot \tan\left(90^{\circ}-\frac{45^{\circ}}{2} \right) \\ a &=& \dfrac{s}{2}\cdot \left(1+\sqrt{2} \right) \\\\ A_{\text{octagon}} &=& \dfrac{a \cdot s}{2} \cdot 8 \\ &=& 4as \\ &=& 4 \left( \dfrac{s}{2}\cdot \left(1+\sqrt{2} \right) \right)s \\ &=& 2 s^2 \left(1+\sqrt{2} \right) \\ \mathbf{ \dfrac{A_{\text{octagon}}} {2} } & \mathbf{=} & \mathbf{s^2 \left(1+\sqrt{2} \right)} \\ \hline \end{array} \)

\(\text{Let $A_{\bigcirc} = \pi r^2$ }\)

\(\begin{array}{|rcll|} \hline \mathbf{A_{\bigcirc}} &=& \mathbf{\dfrac{A_{\text{octagon}}} {2}} \\ \pi r^2 &=& s^2 \left(1+\sqrt{2} \right) \\\\ r^2 &=& \dfrac{s^2 \left(1+\sqrt{2} \right)} {\pi} \\\\ r &=& s \sqrt{ \dfrac{ 1+\sqrt{2} } {\pi} } \quad & | \quad s=4~ cm\\\\ r &=& 4 \sqrt{ \dfrac{ 1+\sqrt{2} } {\pi} } \\\\ r &=& 4 \cdot 0.87662309134 \\ r &=& 3.50649236534 \\ \mathbf{r} & \mathbf{=} & \mathbf{3.5~ cm} \\ \hline \end{array}\)

\(\text{well the easiest way to do this is just list out the 8 digit powers of 2}\\ \left( \begin{array}{cc} 24 & 16777216 \\ 25 & 33554432 \\ 26 & 67108864 \\ \end{array} \right)\)

 \(\text{looks like }33554432=2^{25} \text{ is the winner}\)

\(\text{Split the team up into 2 subteams by height}\\ \text{The tallest 5 is one subteam, the shortest 5 the other}\\ \text{Any member of the tall team can stand in the back}\\ \text{Any member of the short team can stand in front}\\ \text{There are }5! \text{ ways for the 5 tall team members to arrange themselves}\\ \text{There are also }5! \text{ arrangements for the short team members}\\ \text{Thus there are }5! \times 5! = 120^2 = 14400 \text{ different arrangements}\)

Realize that the ratio is inverse. 

The MORE laps in time, the better. 

The LESS time for a mile, the better.

MORE does not equal LESS, so we need to get the reciprocal of LESS to have \(MORE = \dfrac{1}{LESS}\). 

What is the least positive odd integer with exactly nine natural number divisors?

\(\begin{array}{|rcll|} \hline number &=& p^{\color{red}2}q^{\color{red}2} \\ divisors &=& ({\color{red}2}+1)({\color{red}2}+1) = 9 \\ \hline \end{array}\)

The least odd prime numbers are: 3 and 5.

\(\begin{array}{|rcll|} \hline number &=& 3^{\color{red}2}5^{\color{red}2} \\ number &=& 9\cdot 25 \\ \mathbf{number} & \mathbf{=} & \mathbf{225} \\ \hline \end{array} \)

Divisors 225:
1 | 3 | 5 | 9 | 15 | 25 | 45 | 75 | 225 (9 divisors)

Runner A's speed in laps, 17 laps per x. 

Runner B's speed in laps, 6 lags per x. 

The ratio is \(\dfrac{7.5}{z} = \dfrac{17}{6}\). Cross multiplying, we have \(17z = 45 \Rightarrow z = 2.647 \). 2.647 minutes is roughly 159 seconds. 7.5 minutes is 450 seconds.

450 - 159 = \(\boxed{291}\) seconds.

- PM

Let me echo Melody's sentiments.

If you don't know how to solve a problem don't guess at it.  It just confuses and frustrates the posters.

\(\text{This is a subtle problem with an obvious solution of }n=1 \\ \text{and not so obvious solutions of } n=6k,~k \in \mathbb{N}\)

\(\text{A simple bit of algebra shows that }\\ \left(z + \dfrac 1 z\right)^n = 1 \Rightarrow \Large z = e^{\pm i \frac \pi 3}\)

\(\text{On the other hand the roots of }z^n = 1 \text{ are }\\ \Large z = e^{i \frac{2\pi}{n}k},~\normalsize k \in \mathbb{Z}\)

\(\text{So we are looking for }n \text{ such that }\\ 2\pi \dfrac k n = \dfrac \pi 3\\ \dfrac{6k}{n}=1\\ n=6k\)

I am going to try to draw a diagram, but I am not 100% sure if it is the correct shape. Usually, for geometry, we prove things correct, unless it is an indirect proof, then we prove the conclusion wrong and work towards the true conclusion. 

For b), try to follow along with what I did in a). Use the variable "v" to represent the "virtual image's distance from the convex lens".

I'll give you the starting variables. 

Height of windmill = 3cm = height of object

Height of virtual image = 90 cm = h

Lens strength = 10 dpt = focal length 10 cm

- Partial Physician/Mathematician

a) 

Height of arrow = 2cm

Distance from lense = 6cm

Focal length = 2cm

Formula: 1/f = 1/v +1/u               Note: f = focal length, v = image of arrow's distance from lens, u = object's distance from lens

notice that the values are inverted because they are on the opposite side of the lens (opposite side of the arrow)

-1/2 = -1/6 +1/v

1/v = -1/2 + 1/6, which simplifies into 1/v = -1/3

v = -3 cm

This means that the height of the image is upside down, but it is still 3 cm.

(If you know physics, the virtual image behind the convex lens turns upside down if the object is behind the focal point of the convex lens.)

Height of image/Height of object = -v /u = 3/6 = 1/2

Height of image / 2 = 1/2                                Note: height of image (virtual reflection) of arrow = h

Multiply by 2 on both sides to get \(\boxed{h = 1cm}\).

The answer to a) is \(\boxed{A}\).

Hope this helps,

- Partial Physician (Partial Mathematician)

Yes, I think it is isometry. Isometry changes the direction by rotation, reflection, translation, ect. The original size of the shape does not change.

E can also mean a certain power of 10. You might have seen it before in calculators.