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This is a right triangle...its area  =  product of the leg lengths  / 2  = 

15 * 8  / 2   =

120  / 2  =

60  units^2   

The  area  also equals

(1/2) CD  *  AB  ...so

(1/2) CD  * 17   =  60

CD  * 17  =  120

CD  = 120/17

So..by AA concruency......triangle ACB  is similar to triangle  ADC

Therefore    AD /DC  = AC / BC

AD  =  (120/17) * 6 / 15   =  64/17

And in triangle ADC.....AC is the hypotenuse  and  AD and DC  are the legs...so its area  =

(1/2)  ( AD) ( DC)   =

(1/2) (64/17) (120/17)  ≈  13.29 units^2   

Okay, lets work this one through. 

The rectangle with area 3 is sharing its length with the rectangle with area 4. Let this value be \(x\)

The rectangle with area 3 is sharing its width with the rectangle with area 5. Let this value be \(y\)

The rectangle with area 4 is sharing its width with the rectangle with unknown area. Let this value be \(z\)

The rectangle with area 5 is sharing its length with the rectangle with unknown area. Let this value be \(w\)

If we set up and equation for each rectangle, we get the following:

Bottom left rectangle: \(xy = 3\)

Bottom right rectangle: \(wy = 5\)

Top left rectangle: \(xz = 4\)

Top right rectangle: \(wz = ?\)

First off, I am going to solve for the \(x\) in the first equation. We get \(x = \frac{3}{y}\).

Next, I am going to use the second equation to solve for \(y\). We get \(y = \frac{5}{w}\).

I am going to use the second variable we solved for and plug it into the first, like this: \(x = \frac{3}{\frac{5}{w}}\)

If we simplify this eqution, we get: \(x = 3\div \frac{5}{w} = 3(\frac{w}{5}) = \frac{3w}{5}\)

At this point, we know we're getting close to the answer, but we're not done yet.

If we use the value we just got for \(x\) and plug it into the third equation, look what we get:

\(xz = 4 , (\frac{3w}{5})(z) = 4 , (3w)(z) = 20 , wz = \frac{20}{3}\)

Wait...isn't \(wz\) the unknown area?

Therefore, the answer is \(\boxed {\frac{20}{3}}\), or \(\boxed {6.\overline {6}}\)

Okay. Let's work this through step by step. If \(m\angle {BAC} = 70^\circ\), and \(\triangle{ABC} \sim \triangle {PAQ}\), then \(\angle{APQ}\) should also be \(70^\circ\).

If \(\angle{APQ} = 70^\circ , \angle{CPQ} = 110^\circ\).

And since \(\triangle{QPC} \sim \triangle{ABQ} , \angle{AQB} = 110^\circ\).

Also, since \(\angle{AQB} = 110^\circ , \angle{AQC} = 70^\circ\).

Okay, now for the second part.

Since \(\triangle {ABQ} \sim \triangle {QPC} , \angle {PQC} = \angle {BAQ}\).

Let both these angles be \(x\).

This means that \(\angle {QAP} = \angle {AQP}\), since \(\angle {BAC} = 70^\circ\) and \(\angle{AQC} = 70^\circ\), and both \(\angle {QAP}\)  and \(\angle{AQP}\) equal \((70-x)^\circ\).

Since \(\angle {APQ} = 70^\circ\), both  \(\angle {QAP}\)  and \(\angle{AQP}\) are equal to \(55^\circ\).

Now the final step. Now that we have the angle measurements of \(\angle {AQB}\) and \(\angle {AQP}\), we can solve for \(\angle {PQC}\).

\(\angle {AQB} + \angle {AQP} = 165^\circ\).

 \(\angle {PQC} = 180^\circ - 165^\circ = 15^\circ\)

\(\boxed {m\angle {PQC} = 15^\circ}\)

construct a tangent from and external point

EP..I actually think that the guest's answer is  somewhat correct.....here's why

lAlthough  it is true that there are only   6 diferent  combinations of numbers....sequencing is important....for instance

If the correct combination  is   5  4  1   3   6

This isn't the same as   1  3  4  5  6

So...for each combo....we can  permute them in   5! ways

So.....  6  * 5!   =   720

But only one of the selected sequences must be the correct one

So.....  1/ 720   seems to be the  correct answer   ???

To write what he just said in a way that is easier to see,

\((1+i)^{16} = ((i+1)^{8})^{2} = (((i+1)^{4})^{2})^{2} = ((((i+1)^{2})^{2})^{2})^{2}\)

Now if we expand and solve for \((i+1)^{2}\), we get:

\(i^2+2i+1 = -1+2i+1 = 2i\)

Therefore, \(((((i+1)^{2})^{2})^{2})^{2} = (((2i)^2)^2)^2\)

Since \(i = -\sqrt1\)  ,   \((2i)^2 = -4\)

Therefore, \((((2i)^2)^2)^2 = ((-4)^2)^2\)

Evaluating \(((-4)^2)^2\), we get \((16)^2\)

\(16^2=256\), so \(\boxed{256}\) is our answer

Sorry, I can't see that the sum of Sigma is a multiple of 2017, which is a prime number !! Even though they ask to prove it without summing it up, I still summed it up and got 1,986,985,000 which is no multiple of 2017. In order for the sum to be a multiple of 2017, it has to sum up to 1,986,985,023 !!!.

(1+i)^2016 =((1+i)^16)^126 =256^126

(1 - i)^2016=((1-i)^16)^126 =256^126

256^126 - 256^126 =0

Simplify the following:
(i + 1)^16

Compute (1 + i)^16 by repeated squaring. For example a^7 = a a^6 = a (a^3)^2 = a (a a^2)^2.
(1 + i)^16 = ((i + 1)^8)^2 = (((i + 1)^4)^2)^2 = ((((i + 1)^2)^2)^2)^2:
((((i + 1)^2)^2)^2)^2

Expand (i + 1)^2.
(i + 1)^2 = 1 + i + i - 1 = 2 i:
(((2 i)^2)^2)^2

Evaluate (2 i)^2.
(2 i)^2 = -4:
((-4)^2)^2

Evaluate (-4)^2.
(-4)^2 = 16:
16^2

Evaluate 16^2.
=256

 Right so this is my attempt at the question:

From the graph we know that from B A = -b+a and O A  = a

we know that OX = 2a-b, assuming A is in the way if we subtract the value it takes to get to A from O then we get AX= a-b 

as 2a-b - a = a-b.

then we can get the value of BX by adding the "distance" of BA and AX which is 2a-2b.

we know that B, A , X lies on the same staight line because AX (a-b) is a factor of BX (2a-2b).

let me know if im wrong 

The question is right below in the picture

Another look:

Possible combinations         1 2 3 4 5

                                              2 3 4 5 6

                                               1 3 4 5 6

                                                1 2 4 5 6

                                                  1 2 3 5 6    

Only SIX combintaions are possible since you push FIVE of the buttons down at ONCE

                                 So you have a   1 out of 6 chance of getting it correct (as Cphil found)

P = 30 . 97696         x 4 = 123.9078

O=15.9994               x 6 =95.9964

123.9078 / (123.9078 +95.9964)    * 1099.45  = 619.498 gm of P

Thank you!

I think this should go something like this:

Total number of possible combinations, or rather, permutations of 6 numbers is: 6! =720

The number of ways of choosing 5 numbers out of 6 numbers is: 6C5 =6

Therefore, the probability of opening the Lock should be: 6/720 =1/120

CPhill: There is a minor typo in Heron's formula calculation:
sqrt(24 (24 - 21) (24 - 17) (24 - 10))
Sqrt[24 x 3 x 7 x 14]
Sqrt[7,056] =84 - the area of the triangle.

The total number of possible  combinations  = C(6, 5)   =  6

So...the probabilty  is    1 /  6

We can find the area of this triangle using Heron's Formula

s  = [ 10 + 17 + 21  ] / 2  =  48 / 2  = 24

The area  =  √ [ s (s - 10) (s - 17) (s - 21) ]  =  

√ [ 24 (14) (7) (3) ] =

√7056   = 84  units^2

The area   =  1/2  [ sum of triangle's sides ]  *  altitude of each triangle

But....the altitude of each triangle  = the radius  of the incircle

So...we have

84  = (1/2) (48) * radius of incircle

84  = 24 * radius of incircle

84 / 24  = rasius of incircle  = 3.5 units

EDIT TO CORRECT AN ERROR  !!!!

triangle AQB is similar to triangle ZQY

Since  QY  = 24  and BQ  = 12

And   QY + BQ   = 36  = BY

Then   QY  = (2/3)  BY

So... QZ  = (2/3)  AZ   = (2/3) 42  = 28

This has already been asked and answered. If you have a problem with the answer than write that on the original post!

It is rude to repost like this !

No they are not ticked automatically.

We do not have time to check all the answers.

If a question vaguely looks like it might be answered then it gets ticked off.

People can still post on it though. I have to open it before I can tick it off and if someone posts on it afterwards then I can see that easily.

So in a way having a question ticked makes moderators  more  conscious of continuing posts which is a good things. People are more likely to get noticed if they want clarifications.

ok :)

Yes, is it 5 pairs?

Thanks, tertre

If tertre's answer is supposed to be in scientific notation....we have  6.3 x 10³

See, since 5! ends in a zero, all the factorials after that end in a zero! Test it out, yourself. Therefore, 1! ends in 1, and 3! ends in 6. Thus, 1+6+0+0+0+0=7.