Melody

Where is the diagram?

3!*4! = 6*24= 144 ways

$3.05

I have answered on the original question.

the second one is very similar 

You just have to know that 

     \(a^3+b^3=(a+b)(a^2-ab+b^2)\\ \text{rearranging I get}\\ a^3+b^3=(a+b)(a^2+2ab+b^2-3ab)\\ a^3+b^3=(a+b)([a+b]^2-3ab)\\ \text{now check what I have done and do the appropriate substitutions to finish the question.}\)

1)    

\(7x^2+x-5=-4x-9\\ 7x^2+5x+4=0\\\)

The sum of the roots          a+b= -5/7

The product of the roots       ab=4/7

\((a-4)(b-4)\\ =ab-4a-4b+16\\ =ab-4(a+b)+16\\~\\ \text{Substitute and you get}\\ =\frac{4}{7}-4*\frac{-5}{7}+16\\ =\frac{4}{7}+\frac{20}{7}+16\\ =\frac{24}{7}+16\\ =19\frac{3}{7}\)

Hi Juriemagic,

A smart little kid could use trial and error.

Your question is incomplete.

The big hand is on the 11    

The little hand is   11/12 of the way from 8 to 9

So they are both on the left hand side of the clock.

from 9 to 11 is   2/12 *360 = 60 degrees

1/12 of 1/12 * 360 = 2.5  degrees

So I get  62.5 degrees

You need to check.

Firstly I do not think your simplification is correct.  (Unless I have misunderstood what you have wanted to display)

\(e^{-x}+e^{x+e^{-x}}\\ =e^{-x}+e^xe^{e^{-x}}\\ \displaystyle \lim_{x\rightarrow \infty}\;e^{-x}+e^xe^{e^{-x}}\\ =0+e^\infty*1\\ =e^\infty\\ =\infty\)

My presentation is not great. Limits is not my strong suit.  But still I think the logic is correct.

Here is the graph

Start with  5 white and 3 black

1)     P(BB)  \(= \frac{3}{8}*\frac{2}{7}=\frac{6}{56}\)

P(BW) =  \(\frac{3}{8}*\frac{5}{7}=\frac{15}{56}\)

throw back the white ball then P(drawing black)  \(=\frac{2}{7}\)

2)    So  Prob of getting BWB     \(\frac{15}{56}*\frac{2}{7}= \frac{15}{196}\)

3)    The prob of getting  WBB    is the same as   BWB

P(WW)  \(\frac{5}{8}*\frac{4}{7}=\frac{20}{56}\)

Throw both balls back and the prob of then getting BB is  6/56

4) P( WWBB) = (20/56)*(6/56) = \frac{15}{392}

So the prob of ending up with 2 black balls is     \(\frac{15}{56}+\frac{15}{196}+\frac{15}{196}+\frac{15}{392}= \frac{45}{98}\)

sqrt4 x^2 + sqrt x^2=6

You have not used brackets, is this the question?   

\(\sqrt{4 x^2} + \sqrt {x^2}=6\)

\(\sqrt{x^2}=|x| \\SO\\ 2|x|+|x|=6\\ 3*|x|=6\\ |x|=2\\ x=\pm2\)

I have assumed that x is a real number.

You have double counted.

ex:

If candy A,B,C go into the blue bag that is the same as    B,C,A going in the blue bag.

If I have time I will look more later    :)

How many are perpendicular?

How many are horivontal?

how many have a gradient of 1 ?

how many have a gradient of -1 ?

how many have a gradient of 1 ?

how many have a gradient of 1 /2?

how many have a gradient of -1/2 ?

how many have a gradient of 2 ?

how many have a gradient of -2 ?

Add them together  

\(If \quad 625^x=64\qquad then\;find\qquad 125^x\\~\\ 625=5^4\\ 125=5^3\\ 64=4^3=2^6\\~\\ 625^x=64\\ 5^{3x\cdot \frac{4}{3}}=64\\ (125^x)^{\frac{4}{3}}=4^3\\ (125^x)=4^{3 \cdot \frac{3}{4}}\\ 125^x=4^{\frac{9}{4}}\\ 125^x=4^2*4^{1/4}\\ 125^x=4^2*2^{1/2}\\ 125^x=16\sqrt2 \)

NOTE:  I could easily have made a careless error.

LaTex

If \quad 625^x=64\qquad then\;find\qquad 125^x\\~\\
625=5^4\\
125=5^3\\
64=4^3=2^6\\~\\
625^x=64\\
5^{3x\cdot \frac{4}{3}}=64\\
(125^x)^{\frac{4}{3}}=4^3\\
(125^x)=4^{3 \cdot \frac{3}{4}}\\
125^x=4^{\frac{9}{4}}\\
125^x=4^2*4^{1/4}\\
125^x=4^2*2^{1/2}\\
125^x=16\sqrt2

a=10, b=6, c=9, d=7

If you care about the logic then just ask.

Prob of 3 s in three bats is    0.323^3  = approx 0.0337 = approx  3.4% = approx  0.03

Put one adult anywhere.

then there are 2 posibilities for how to sit the other 2 adults 

Tthen 3! = 6 ways to seat the chiildren.

So that is  2*6=12 ways