All Answers 359332 Answers

y=\(\sqrt{3-\sqrt{x}} \)

\(y'=\ ?\)

\(\begin{array}{|rcll|} \hline y &=& \sqrt{3-\sqrt{x}} \\ y^2 &=& 3-\sqrt{x} \quad | \quad \text{derivate both sides} \\\\ 2yy' &=& -\dfrac{d\ ( \sqrt{x})}{dx} \\\\ 2yy' &=& -\dfrac{d\ (x^{\frac12})}{dx} \\\\ 2yy' &=& -\dfrac{1}{2} x^{-\frac{1}{2}} \\\\ 2yy' &=& -\dfrac{1}{2x^{\frac{1}{2}}} \\\\ 2yy' &=& -\dfrac{1}{2\sqrt{x}} \\\\ y' &=& -\dfrac{1}{2\sqrt{x}2y} \\\\ y' &=& -\dfrac{1}{4\sqrt{x} y} \\\\ \mathbf{y'} &=& \mathbf{-\dfrac{1}{4\sqrt{x} \sqrt{3-\sqrt{x}}}} \\ \hline \end{array}\)

2. How many ways are there to put 6 balls in 3 boxes if the balls are not distinguishable but the boxes are?

Again....assuming that we can have empty boxes.....the number of ways of distributing k  identical balls into n  distinguishable boxes is given by :

C ( k +n - 1 , n - 1)   =  C (6 +3 - 1, 3 - 1 )  = C (8,2)  =  28 ways

1. How many ways are there to put 6 balls in 3 boxes if the balls are not distinguishable and neither are the boxes?

Assuming that  we can have empty boxes.....this boils down to the number of ways that we partition  6  identical balls into 3 identical boxes

These are

6, 0, 0

5, 1, 0

4, 2, 0

4, 1, 1

3, 3, 0

3, 2, 1

2, 2, 2

So......7 ways

 Rate x time = distance

Rate = Speed + wind    or speed - wind

speed-wind    *  2   = 840

Speed + wind  * 1.45 = 840          Two equations    with    two unknowns  

From first equation   W+420 =S      Sub in to the second equation

(W+420+W)*1.45 = 840

W= 30 mph

Sub this result in to one of the equations to solve for  S

S-30       * 2 =  840         S = 450  mph

The numbers \(20\), \(99\), and \(101\) form a Pythagorean triple.
A right triangle has heights \(x\), \(\dfrac{20}{101}\), and \(\dfrac{99}{101}\),

where \(x\) is the shortest height.
What is \(x\)?

Formula:

\(\begin{array}{|rcll|} \hline x &=& \dfrac{\dfrac{20}{101}\cdot \dfrac{99}{101}}{\dfrac{101}{101}} \\\\ &=& \dfrac{\dfrac{20\cdot 99}{101^2}}{1} \\\\ &=& \dfrac{20\cdot 99}{101^2} \\\\ &=& \dfrac{1980}{10201} \\\\ \mathbf{x} &=& \mathbf{0.19409861778} \\ \hline \end{array}\)

Indeed, 3 is correct, read it wrong

Convert \(999_{16}\) to decimal

\(\text{Let ${\color{red}9}{\color{blue}9}{\color{green}9} $}\)

\(\begin{array}{|rcll|} \hline {\color{red}9}\cdot 16 &=& 144 \\ 144 + {\color{blue}9} &=& 153 \\ 153 \cdot 16 &=& 2448 \\ 2448 + {\color{green}9} &=& 2457 \\ \hline \end{array} \)

\(\mathbf{999_{16} = 2457_{10} }\)

Note that quadrilaterals EAHD and  BFCG  must  be  3/4 of the area of the square

But the area of  triangle EAH and  triangle FCG  are 1/2 the area of each of the above quadrilaterals

So.......their combined area  must be   (1/2) (3/4)  =  3/8  units^2

So [ FEGH]  +[ EAH ]  +[ FCG]   = [FACH]   =  1/4  + 3/8   =   2/8 + 3/8  =   5/8 units^2

Convert 35 from base 10 to base 2

\(\begin{array}{|rclcl|} \hline 35 &=& 2\cdot {\color{blue}17} &+& \color{red} 1 \\ {\color{blue}17} &=& 2\cdot {\color{blue}8} &+& \color{red} 1 \\ {\color{blue}8} &=& 2\cdot {\color{blue}4} &+& \color{red} 0 \\ {\color{blue}4} &=& 2\cdot {\color{blue}2} &+& \color{red} 0 \\ {\color{blue}2} &=& 2\cdot {\color{blue}1} &+& \color{red} 0 \\ {\color{blue}1} &=& 2\cdot {\color{blue}0} &+& \color{red} 1 \\ \hline \end{array}\)

\(\mathbf{35_{10} = {\color{red}100011}_2}\)

Then read the remainders from bottom to top to get  100011

So     35₁₀   =   100011₂

Let

\(z=-8+15i \text{ and }w=6-8i.\)

Compute

\(\dfrac{z\overline z}{w\overline w}\),

where the bar represents the complex conjugate.

\(\begin{array}{|rcl|rcl|} \hline z\overline z &=& |z|^2 & w\overline w &=& |w|^2 \\ \hline z &=&-8+15i & w&=&6-8i \\ |z| &=& \sqrt{8^2+15^2} & |w| &=& \sqrt{6^2+8^2} \\ |z|^2 &=& 8^2+15^2 & |w|^2 &=& 6^2+8^2 \\ \hline \mathbf{\dfrac{z\overline z}{w\overline w}} &=& \dfrac{|z|^2}{|w|^2} \\ &=& \dfrac{8^2+15^2}{6^2+8^2 } \\ &=& \mathbf{ \dfrac{289}{100} } \\ \hline \end{array} \)

Here's the way I think of it if it helps:

999₁₆   =   9(16⁰)   +   9(16¹)  +  9(16²)

999₁₆   =   9(1)   +   9(16)  +  9(256)

999₁₆   =   9  +  144  +  2304

999₁₆   =   2457     that is in base 10

--

Ok, awesome!!!

sin( angle )  =  opposite / hypotenuse

sin( C )  =  \(\frac{20}{101}\) / \(\frac{101}{101}\)

sin( C  )  =  \(\frac{20}{101}\)

Also...

sin( C )  =  x / \(\frac{99}{101}\)

sin( C )  =  \(\frac{101}{99}x\)

So we can equate both expressions of  sin( C )  and solve for  x

\(\frac{20}{101}\ =\ \frac{101}{99}x\\~\\ \frac{20}{101}\cdot\frac{99}{101}\ =\ x\\~\\ \frac{1980}{10201}\ =\ x \)_

4. What fraction of all the 10-digit numbers with distinct digits have the property that the sum of every pair of neighboring digits is odd?

Here's my best attempt   :

We either have

Even Odd  Even Odd  Even Odd Even Odd  Even Odd        or

Odd Even  Odd Even Odd Even  Odd Even  Odd Even

In  the first case  we have the following choices for each position....note that the first position cannot be 0, so we only have 4 evens to choose from

4 * 5 * 4 * 4 * 3 * 3 * 2 * 2 * 1 * 1  =    5! * 4 * 4!  =  11520  integers

In the second case we have

5 * 5 * 4 * 5 * 3 * 3 * 2 * 2 * 1 * 1  =    5! * 5!  = 14400 integers 

So    11520 + 14400  = 25920

And we have   10^10  - 10^9    =  9000000000   ten digit integers

So....the fraction is      25920 / 9000000000

Nice job, hectictar   !!!

1. Each of the digits 1,2,3,4,5,6 is used exactly once in forming the 3-digit integers X and Y. How many possible values of X+Y are there if |X-Y|=111?

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To help see this...here is the beginning of the list of pairs:

123, 234

124, 235

125, 236

132, 243

134, 245

135, 246

142, 253

143, 254

.

.

.

Don't know..

5.Mort walks each day for exercise and likes to vary his path. He starts each day at position 1 on the circle shown, then walks in a straight line to another numbered point on the circle, and continues, before returning to his starting point. If everyday Mort visits at least one point numbered 2 -- 6 but doesn't visit the same point twice (except for the starting and ending point), how many different possible paths can he take?

Since we can draw  a (straight-line) chord between  any two points on a circle.....the number of possible paths is just al lthe permutations  of the set   { 1,2,3,4,5,6,1 }    with 1  as the first and last elements

So....this  is just   5!  =   120 possible paths

OK....we agree  !!!!

Mmmm....I think it might have to do with the fact that we have " different" chairs

Maybe this is the approach, but I'm not sure

Call the chair   colors    

Red     Blue    Green   Yellow  White 

The boys can  occupy the chairs in the following ways

(R B)  

(B G)

(G Y)

(Y W)

(W R) 

And for each of these    the boys can be seated in two ways   =  5 * 2  = 10 ways

And for each of these ways, the girls can be seated in 3!  = 6 ways

So.....maybe   10 * 3!  =  10 * 6   = 60  is your answer ????