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Use difference of squares formula. \((a-b)^3=(a-b)(a^2+ab+b^2)\)

We can solve this problem by using casework and counting the number of possibilities for each case.

Case 1: The number has one digit that is equal to 1 and the other digits are equal to 2. There are 8 ways to choose the position of the digit that is equal to 1 (it can be any of the 8 digits). Once we have chosen the position of the digit that is equal to 1, there are 2 possibilities for each of the other digits, for a total of $2^7$ possibilities.

Case 2: The number has two digits that are equal to 1 and the other digits are equal to 2. There are $\binom{8}{2} = 28$ ways to choose the positions of the digits that are equal to 1. Once we have chosen the positions of the digits that are equal to 1, there are 2 possibilities for each of the other digits, for a total of $2^6$ possibilities.

Case 3: The number has three digits that are equal to 1 and the other digits are equal to 2. There are $\binom{8}{3} = 56$ ways to choose the positions of the digits that are equal to 1. Once we have chosen the positions of the digits that are equal to 1, there are 2 possibilities for each of the other digits, for a total of $2^5$ possibilities.

Case 4: The number has four digits that are equal to 1 and the other digits are equal to 2. There are $\binom{8}{4} = 70$ ways to choose the positions of the digits that are equal to 1. Once we have chosen the positions of the digits that are equal to 1, there are 2 possibilities for each of the other digits, for a total of $2^4$ possibilities.

Case 5: The number has five digits that are equal to 1 and the other digits are equal to 2. There are $\binom{8}{5} = 56$ ways to choose the positions of the digits that are equal to 1. Once we have chosen the positions of the digits that are equal to 1, there are 2 possibilities for each of the other digits, for a total of $2^3$ possibilities.

Case 6: The number has six digits that are equal to 1 and the other digits are equal to 2. There are $\binom{8}{6} = 28$ ways to choose the positions of the digits that are equal to 1. Once we have chosen the positions of the digits that are equal to 1, there are 2 possibilities for each of the other digits, for a total of $2^2$ possibilities.

Case 7: The number has seven digits that are equal to 1 and the other digit is equal to 2. There are 8 ways to choose the position of the digit that is equal to 2. Once we have chosen the position of the digit that is equal to 2, there are 2 possibilities for each of the other digits, for a total of $2^7$ possibilities.

Case 8: The number has eight digits that are equal to 2. There is only one possibility in this case.

Thus, the total number of positive integers that satisfy both conditions is:

2^7 + 2^6 + 2^5 + 2^4 + 2^3 + 2^2 + 2^7 + 1 = 381.

Zero chance   ....if there are  three pieces and they are all less than two inches....that is less than the original 6 inch total length. 

3/8   is notation for    3 ÷ 8     if you use tour calculator , you will find = .375 

Not sure what the question is ....but 

x^2 -12   =   ( x- sqrt12)(x+sqrt 12)            

Are you familiar with the order guidelines used complete calculations of equations called PEDMAS ? 
parentheses  first

exponents    next 

division/multiplication next 

addition/subtraction next

  using these rules 

-2 + 1^2       do the exponent firist     -2 +1     now do the addition   = -1 

(-2+1)^2      do the parentheses first   -1 ^2    now do the exponent   = 1 

The sum of the numbers in each circle is 18.

This simplifies to 6x^2 + 4x - 8.

Let $F$ be the foot of the altitude from $H$ to $BC$. Since $H$ lies on altitude $AD$, we have $\angle AHB = 90^\circ$, which means $\angle BHF = 90^\circ - \angle AHB/2 = 56^\circ$. Also, since $H$ lies on altitude $BE$, we have $\angle AHB = 90^\circ$, which means $\angle AHB = 180^\circ - \angle BAC$. Thus, $\angle BAC = 180^\circ - 2\angle BAH = 124^\circ$.

Now, we use the fact that the sum of the angles in a triangle is $180^\circ$ to find $\angle HCA$. We have:

\begin{align*}
\angle HCA &= 180^\circ - \angle ACH - \angle AHC \
&= 180^\circ - \angle ACH - (180^\circ - \angle BHA) \
&= \angle BHA - \angle ACH \
&= (180^\circ - \angle BAC) - \angle ACH \
&= 56^\circ - \angle ACH.
\end{align*}

Thus, we need to find $\angle ACH$. Since $AF$ is the altitude of triangle $ABH$, we have $\angle ABF = 90^\circ$. Since $\angle BAH = 28^\circ$, we have $\angle BAF = \angle BAH + \angle HAF = 28^\circ + 90^\circ - \angle AHB/2 = 46^\circ$.

Using the fact that the sum of the angles in a triangle is $180^\circ$ in triangle $ABC$, we have:

\begin{align*}
\angle ACH &= 180^\circ - \angle A - \angle CHA \
&= 180^\circ - \angle A - (180^\circ - \angle BHF - \angle BHA) \
&= \angle BHA - \angle A - \angle BHF \
&= (180^\circ - \angle BAC) - \angle A - \angle BHF \
&= 56^\circ - \angle A - \angle BHF.
\end{align*}

Thus, we need to find $\angle A$ and $\angle BHF$. Since $AF$ is the altitude of triangle $ABH$, we have $\angle AFB = 90^\circ$, so $\angle BAF + \angle ABF = 46^\circ + 90^\circ = 136^\circ$. Since $\angle BAC = 124^\circ$, we have $\angle BCA = \angle BAC/2 = 62^\circ$. Thus, $\angle A = 180^\circ - \angle BAC - \angle BCA = 180^\circ - 124^\circ - 62^\circ = 14^\circ$.

To find $\angle BHF$, note that $\angle BHF + \angle BHA + \angle AHF = 180^\circ$. Since $\angle BHA = 128^\circ$ and $\angle AHF = 90^\circ - \angle BAF = 44^\circ$, we have $\angle BHF = 180^\circ - \angle BHA - \angle AHF = 8^\circ$.

Therefore, $\angle ACH = 56^\circ - \angle A - \angle BHF = 56^\circ - 14^\circ - 8^\circ = 34^\circ$.

Take anyone of the 6 kids:

He/she has 6 choices. The sibling has 5 - 1 = 4 choices.

The 4th kid has the remaining 4 choices. The sibling has 3 - 1=2 choices

The last pair has the remaining 2 choices.

Total seating =6  x  4  x  4  x  2  x  2 =384 ways of seating the 3 pairs of siblings.

This question has been asked a lot ot times recently.

there are probably a number of different answers given.

I let the children be     A1    A2    B1    B2    C1    C2

If there were no restrictions there would be 6! ways to seat them

How many ways can they NOT be seated.

We cannot have all the pairs under each other so  that would be

6 places to put A1    then 1 place to put A2

4 places to put B1   then 1 place to put  B2

2 places to put  C1  then 1 place to put C2

6*4*2 = 48

We cannot have one pair vertically alligned with the others not so

Say the As are vertially alligned but not B or C

6 places to put A1, 1 place to put A2

4 places to put B1, 2 places to put B2

2 places to put C1 and 1 place to put C2

6*1*4*2*2*1 = 96 ways

BUT ic could have been A or B or C that are vertically aligns so we have to multiply by 3

96*3=288

So I think the answer is   

6! - (48+288) = 6! - 336 = 720 - 336  =  384  ways

The triangles XWZ and XYW (in that order) are similar.

( The angles XWZ and XYW are both equal to the angle between the tangent and the line XW,

XWZ because of the parallel lines, and XYW by the circle property, angle in the opposite segment. 

Angle WXY is common to both triangles.)

So,

XZ/XW = XW/XY,

XZ = XW^2/XY = 6^2/14 = 36/14 = 18/7.

Then,

YZ = 14 -18/7 = 80/7.

The 66/7 mentioned at the bottom of the question refers to an earlier posting. 

A version of this question, it had XW = 8 rather than 6, was posted July(?)/ August 2020.

(1,5), (5,1), (1,7), (7,1), (5,7), (7,5) ==6 - any of these 6 pairs included in 10! permutations must be removed.

10!  -  [9! * 6]  +  [8! * 6] ==1,693,440 - total number of ways or permutations.

1122444 ( 210 ) , 1222444 ( 140 ) , 1223444 ( 420 ) , 1224444 ( 105 ) , 1224445 ( 420 ) , 1224446 ( 420 ) , >>Total combinations = 6

>>Total permutations(totals in brackets) = 1,715 permutations - number of possible sequences of rolls.

Let O be the center of the circumcircle of triangle WXY, and let T be the point where the tangent at X intersects the circumcircle.

Since the tangent at X is perpendicular to the radius OX, we have OT ⊥ XT. Also, since WX is a chord of the circumcircle, we have OT ⊥ WX. Therefore, OT is perpendicular to the line through W that is parallel to the tangent, and so the line through W that is parallel to the tangent passes through the midpoint M of OT.

Since M is the midpoint of OT, we have OM = MT. Also, since OX is a radius of the circumcircle, we have OM = OX. Therefore, MT = OX = 6.

Let P be the point where the line through W that is parallel to the tangent intersects XT. Then, since the line through W is parallel to the tangent, we have ∠XWP = ∠XTX = 90°.

Since ∠XWP = 90°, we have WP ⊥ WX. Also, since MT = 6, we have PT = MT = 6. Therefore, by the Pythagorean Theorem, we have WP = √(XT^2 - PT^2) = √(XT^2 - 36).

Since WP ⊥ WX, we have ∠WPX = 90°. Therefore, by the Pythagorean Theorem, we have WX^2 + XP^2 = WP^2. Substituting WP = √(XT^2 - 36), we get:

6^2 + XP^2 = (XT^2 - 36) XP^2 = XT^2 - 36 - 6^2 XP^2 = XT^2 - 72

Since T is on the circumcircle, we have OT = r, where r is the radius of the circumcircle. Also, since OT ⊥ XT, we have OP = PT = 6. Therefore, by the Pythagorean Theorem, we have OT^2 = OP^2 + PT^2 = 6^2 + 6^2 = 72.

Now, let Q be the point where the line through W that is parallel to the tangent intersects the line through T that is perpendicular to WX. Then, since ∠XWP = 90° and ∠XTW = 90°, we have ∠WXT = ∠XPQ. Therefore, triangles WXT and XPQ are similar, so we have:

XP/XQ = WX/WT XP/(XQ + WX) = WX/OT XP/(XQ + 6) = 6/√72 XP/(XQ + 6) = √2/2

Solving for XP, we get:

XP = (√2/2)(XQ + 6)

Substituting into the previous equation, we get:

(√2/2)(XQ + 6)^2 = XT^2 - 72

Expanding and simplifying, we get:

XQ^2 + 12XQ - 2XT^2 + 144 = 0

Substituting XT = 6√2, we get:

XQ^2 + 12XQ - 72 = 0

Using the quadratic formula, we get:

XQ = (-12 ± √(12^2 + 4(72)))/2 XQ = (-12 ± 6√5)/2

Since XQ > 0, we have:

XQ = -6 + 3√5

Finally, using the Pythagorean Theorem on YTZ, we get

YZ = 8 + 2√5

I guess you could just use long division. You could also say that 1/8 is 0.125 so you could multiply 0.125 times 3. If you have any questions, just ask :)

\(\sqrt{36x^{36}}=\sqrt{6^2x^{18\times2}}=\sqrt{6^2(x^{18})^2}=6x^{18}\)