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Are these questions in a book or something?  I answered this question

two months ago at https://web2.0calc.com/questions/geometry_68515

3 and 9 o'clock

Similar triangles....CD is 3 times bigger than BC   so the larger triangle is dilated by factor of 3 ...each side is 3 times bigger than the smaller

   area of larger = 3 * 3  * 6  = 54 cm²

Both answers (Dragan's and heureka's) are correct!!!

[XYZ] = 340.7076581   (!!!)

forgot to add this

$\binom{12}{4} + \binom{11}{4} + \binom{19}{4} = \boxed{4701}$

I don't think these numbers are possible    if 48 take spanish only , then the number who take chinese only has to be 53    which is over the 97 student limit .

(x-4)^2 + (21- -7)^2  = (10 sqrt3)^2

x^2 - 8x + 800 = 300

x^2 - 8x + 500 = 0         quadratic formula or factoring shows x = 4 +- 22i      summed = 8

It will be the same disatnce from the vertex BUT on the OTHER side

6,0        x coordinate = 6

wrong

By stars and bars, the number of ways is 19!/(4!*7!*8!) = 24942060.

Numbers that cause the DENOMINATOR to = 0 are NOT in the domain

0   +1     +2    = 3

Find CB via pythagorean theorem to be = sqrt5

TanB = CD/CB = 3 / sqrt5

Using angle bisector theorem, BC would be 16/3. Using the angle Bisector theorem again, 10 / AZ = (16/3)  / BZ. Can you take it from here?

Need to 'massage' this around to vertex form of a parabola   ( a sideways parabola due to x = y^2 form)   x = a(y-k)^2 + h

x - y^2 + 8y = 13 + 6y + y^2

x = 2y^2 -2y    + 13        complete the square for 'y'

x = 2 ( y^2-1)   + 13

x = 2 ( y-1/2)^2   - 1/2 + 13

x = 2 (y-1/2)^2 + 12.5

           vertex is   (12.5 , .5)

the rest of the problem reads: For example, we could obtain 5 x 55 x 55, 555 x 55, or 55555, but not 5 x 5 or 2525. I'm confused too

Not sure what you are asking, but maybe

5 x 5

5 x 55

5 x 555

5 x 5555

55 x 55

55 x 555

For ODD 3 digit integers

   4 choices for first digit

   4 choices for second digit

   two choices for third digit ( 1 or 3) =   4 x 4 x 2 = 32   three digit ODD numbers

Solutions:

(1) Calculate the total number of 13-card hands when face-cards, aces, and tens are removed.

Divide this by the total number of 13-card hands from the complete deck

 nCr(32, 13) / nCr(52, 13) = 0.0005470333581834

This corresponds to odds of about (1) in (1829) hands.

(2)  This is a conditional probability. Given that one player has a Yarborough, what is the probability that a second player also has a Yarborough?

Calculate the total number of 13-card hands after an additional 13 cards are removed from the reduced deck. Divide this by the total number of 13-card hands from the complete deck.

nCr(32, 13) / nCr(52, 13) = 0.0000000427266467.

The over all probability of two players being dealt a Yarborough in a single round is

(0.0005470333581834) * (0.0000000427266467) = 0.00000000002337290102821668560478

This corresponds to odds of about (1) in (42 784 590 531) rounds,

GA

If

\(f(x) = \dfrac{a}{x + 2}\),

solve for the value of \(a\) such that \(f(0) = f^{-1}(-3a)\).

My attempt:
\(\begin{array}{|rcll|} \hline f(x) &=& \dfrac{a}{x + 2} \\\\ f(0) &=& \dfrac{a}{0 + 2} \\\\ \mathbf{f(0)} &=& \mathbf{\dfrac{a}{2}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline f(x) &=& \dfrac{a}{x+2} \\\\ x &=& \dfrac{a}{f^{-1}(x)+2} \\\\ x\Big(f^{-1}(x)+2\Big) &=& a \\\\ xf^{-1}(x)+2x &=& a \\\\ xf^{-1}(x) &=& a-2x \\\\ f^{-1}(x) &=& \dfrac{a-2x}{x} \quad | \quad x =-3a\\\\ f^{-1}(-3a) &=& \dfrac{a-2(-3a)}{-3a} \\\\ f^{-1}(-3a) &=& \dfrac{a+6a}{-3a} \\\\ f^{-1}(-3a) &=& \dfrac{7a}{-3a} \\\\ \mathbf{f^{-1}(-3a)} &=& \mathbf{-\dfrac{7}{3}} \\\\ \hline f(0) &=& f^{-1}(-3a) \\\\ \dfrac{a}{2} &=& -\dfrac{7}{3} \\\\ \mathbf{a} &=& \mathbf{-\dfrac{14}{3}} \\ \hline \end{array}\)

Melody: They want "How many three digit ODD numbers" :

111 ( 1 ) ,  113 ( 3 ) , 123 ( 6 ) , 133 ( 3 ) , 223 ( 3 ) , 233 ( 3 ) , 333 ( 1 ) >>Total combinations = 7
>>Total permutations =20 [Numbers in parentheses]

If

\(h(y) = \dfrac{1 + y}{2 - y}\),

then what is the value of \(h^{-1}(15)\)?

My attempt:

\(\begin{array}{|rcll|} \hline h^{-1}\Big(h(y)\Big) &=& y \\\\ &&\boxed{h(y) = 15 \\ \frac{1+y}{2-y}=15\\ 1+y = 15(2-y) \\ 1+y=30-15y\\ 16y = 29 \\ \mathbf{y=\frac{29}{16}} } \\\\ \mathbf{h^{-1}(15)} &=& \mathbf{\dfrac{29}{16}} \\ \hline \end{array}\)

If

\(a + b = 7\) and \(a^3 + b^3 = 42 + ab\),

what is the value of the sum \(\dfrac{1}{a} + \dfrac{1}{b}\)?

\(\begin{array}{|rcll|} \hline \left(a + b\right)^3 =&=& a^3+3a^2b+3ab^2+b^3 \\ \left(a + b\right)^3 =&=& a^3+b^3+3ab(a+b) \\ 7^3 =&=& 42+ab+3ab*7 \\ 343 =&=& 42+22ab \\ 22ab=&=& 343-42\\ ab &=& \dfrac{301}{22} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \dfrac{1}{a} + \dfrac{1}{b} &=& \dfrac{a+b}{ab} \\\\ \dfrac{1}{a} + \dfrac{1}{b} &=& \dfrac{7}{\frac{301}{22}} \\\\ \dfrac{1}{a} + \dfrac{1}{b} &=& \dfrac{7*22}{301} \\\\ \dfrac{1}{a} + \dfrac{1}{b} &=& \dfrac{22}{43} \\ \hline \end{array}\)

How many zeros are at the end of \((100!)(200!)(300!)(400!)\) when multiplied out?

\(\begin{array}{|rcll|} \hline \text{zeros} &=& \small{ \lfloor\frac{100}{5}\rfloor +\lfloor\frac{100}{25}\rfloor +\lfloor\frac{200}{5}\rfloor +\lfloor\frac{200}{25}\rfloor +\lfloor\frac{200}{125}\rfloor +\lfloor\frac{300}{5}\rfloor +\lfloor\frac{300}{25}\rfloor +\lfloor\frac{300}{125}\rfloor +\lfloor\frac{400}{5}\rfloor +\lfloor\frac{400}{25}\rfloor +\lfloor\frac{400}{125}\rfloor } \\\\ &=& 20+4+40+8+1+60+12+2+80+16+3 \\ \mathbf{\text{zeros}} &=& \mathbf{246} \\ \hline \end{array}\)