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Try  7 and  -1

(1/2) sin(2θ) + (1/4)sin(4θ) =

(1/2)*2sinθcosθ + (1/4)sin(2θ + 2θ) =

sinθcosθ + (1/4)[ sin2θcos2θ +  sin2θcos2θ ] =

sinθcosθ  + (1/4) [ (2)[sin2θ][cos2θ] ] =

sinθcosθ  + (1/4) (2) [ [sin2θ][cos2θ] ] =

sinθcosθ + (1/2)[2sinθcosθ]* [2cos^2θ - 1] =

sinθcosθ + sinθcosθ * [ 2cos^2θ - 1 ] =

sinθcosθ + sinθcosθ *  2cos^2θ  - sinθcosθ =

 sinθcosθ *  2cos^2θ  =

2 sinθcosθ *  cos^2θ    

sin(2θ) * cos^2θ      and replacing θ  with pi * x .....we have

sin(2* pi *x ) * cos^2(pi * x)

And that's it, GL  ....!!!!

V = (1\3)*3.14*(3^2)*8 =

(1/3) * 3.14 * 9 * 8 =

(1/3) * 3.14 * 72 =

24 * 3.14 =

75.36 units³

b                      

2800 / 5  = 560   = 5.6 x 10²

I'm assuming this is

3^352^(log10^(3^(1/4))

3^(1/4) = 1.3160740129524925  ... so we have

3^352^log10^1.3160740129524925 =

3^352^1.3160740129524925 =

3^463.25805255927736

In terms of base 10 .... 3 = 0.4771212547196624

So, we have

(10^0.4771212547196624)^463.25805255927736  =

10^221.030263296069725261145127163264 =

10^221 * 10^0.030263296069725261145127163264 = 

10^221 * 1.072   {approximately } 

Or, in scientifiic notation

1.072 x  10^221  = 1072 followed by 218 zeroes..... (more or less)

Not sure how to give CPhill's answer a vote or thumbs up, but thank you, very helpful! 

Here's a picture of a  regular octahedron :

There are 6 vertices and the number of ways of pairing any two of them is C(6,2)  = 15

But notice that only 12 pairs of them form edges. {The top and bottom vertices aren't connected to each other and neither are the two opposite vertices on the "sides." }

So, the probability that any two form an edge is  12/15 = 4/5

We can use the sine here

sin 13 = 400 / d     where d is the distance on the hypotenuse it has traveled

Rearrange..and we have

d = 400/sin 13 = about 1800 ft (rounded)

This is impossible.....two sum of any sides of any triangle is always greater than the remaining side.

But  99 + 1212  <  1515

The number of total possibilities is C(10,4 ) = 210

But the sum of any four of the first ten prime numbers is odd only if 2 is included.

So, this means that, out of the other nine, we want to choose any three of them.

So C(9,3)  = 84

So...the probability that the sum of any four of the first ten prime numbers is odd is given by ;

84 / 210 =  28/70 = 14/35

the temperature rose 3 Fahrenheit in the afternoon and fell in 3F in the evening. what was the total change in temperature at the end of the day? 

a. -3 fahrenheit

b.0 fahrenheit

c. 3 fahrenheit

d. 6 fahrenheit

which one? 

This site calculator will solve symbolic math. You can use it to test your solutions.

This returns the solution for the question above.

$$\underset{\,\,\,\,{\textcolor[rgb]{0.66,0.66,0.66}{\rightarrow {\mathtt{c1, d, c2, c3}}}}}{{solve}}{\left(\begin{array}{l}{\mathtt{c1}}={\frac{{\mathtt{\pi}}{\mathtt{\,\times\,}}{{\mathtt{d}}}^{{\mathtt{3}}}}{{\mathtt{6}}}}\\
{\mathtt{c2}}={\frac{{\mathtt{\pi}}{\mathtt{\,\times\,}}{{\mathtt{d}}}^{{\mathtt{3}}}}{{\mathtt{4}}}}\\
{\mathtt{c3}}=\left({\mathtt{1}}{\mathtt{\,-\,}}\left({\frac{{\mathtt{c1}}}{{\mathtt{c2}}}}\right)\right)\end{array}\right)} \Rightarrow \left\{ \begin{array}{l}{\mathtt{c1}} = {\mathtt{0}}\\
{\mathtt{d}} = {\mathtt{0}}\\
{\mathtt{c2}} = {\mathtt{0}}\\
{\mathtt{c3}} = {\mathtt{r12}}\\
{\mathtt{c1}} = {\frac{{{\mathtt{r13}}}^{{\mathtt{3}}}{\mathtt{\,\times\,}}{\mathtt{\pi}}}{{\mathtt{6}}}}\\
{\mathtt{d}} = {\mathtt{r13}}\\
{\mathtt{c2}} = {\frac{{{\mathtt{r13}}}^{{\mathtt{3}}}{\mathtt{\,\times\,}}{\mathtt{\pi}}}{{\mathtt{4}}}}\\
{\mathtt{c3}} = {\frac{{\mathtt{1}}}{{\mathtt{3}}}}\\
\end{array} \right\}$$

The result is the same as CPhill’s solution.

Note that all the Cs & Ds are not affecting the result. This calculator has a high resistance to CDD

Here’s another one

$$\underset{\,\,\,\,{\textcolor[rgb]{0.66,0.66,0.66}{\rightarrow {\mathtt{c1, d, c2, x}}}}}{{solve}}{\left(\begin{array}{l}{\mathtt{c1}}={\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{\pi}}{\mathtt{\,\times\,}}{\mathtt{0.5}}{d}\\
{\mathtt{c2}}={\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{\pi}}{\mathtt{\,\times\,}}\left({\mathtt{0.5}}{d}{\mathtt{\,\small\textbf+\,}}{\mathtt{x}}\right)\\
{\mathtt{c2}}={\mathtt{c1}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\end{array}\right)} \Rightarrow \left\{ \begin{array}{l}{\mathtt{c1}} = {\mathtt{r9}}{\mathtt{\,\times\,}}{\mathtt{\pi}}\\
{\mathtt{d}} = {\mathtt{r9}}\\
{\mathtt{c2}} = {\mathtt{r9}}{\mathtt{\,\times\,}}{\mathtt{\pi}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\\
{\mathtt{x}} = {\frac{{\mathtt{1}}}{\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{\pi}}\right)}}\\
\end{array} \right\}$$ 

^ - - - Does anyone know what it is for? It’s easy.

This one is totally symbolic.

$$\underset{\,\,\,\,{\textcolor[rgb]{0.66,0.66,0.66}{\rightarrow {\mathtt{c2}}}}}{{solve}}{\left(\begin{array}{l}{\frac{{\mathtt{c2}}{\mathtt{\,\times\,}}{\mathtt{d2}}}{{\mathtt{t2}}}}={\frac{{\mathtt{c1}}{\mathtt{\,\times\,}}{\mathtt{d1}}}{{\mathtt{t1}}}}\end{array}\right)} \Rightarrow {\mathtt{c2}} = {\frac{{\mathtt{c1}}{\mathtt{\,\times\,}}{\mathtt{d1}}{\mathtt{\,\times\,}}{\mathtt{t2}}}{\left({\mathtt{d2}}{\mathtt{\,\times\,}}{\mathtt{t1}}\right)}}$$

1.4^3 = 2.744 (1.4*1.4*1.4) 

Just divide 43.25 by 12.5......

f(x ) = x^3(x+2)(x-4)

We could use the product/power rules, but I find just expanding this and using the power rule by itself is easier.

So we have

f(x) = x^3(x^2 - 2x - 8)  =  x^5 - 2x^4 - 8x^3

And 

f ' (x) =  5x^4 - 8x^3 - 24x^2

The volume of the cylinder is

V = pi * r^2 * h   where r is the radius of the cylinder and h is the height.

And the radius of the ball = the radius of the cylinder. 

But since the ball touches the sides, top and bottom, the height  of the cylinder must be 2r.

So the volume of the cylinder is

V = pi * r^2 * 2r  =  2pi *r^3

And the volume of the ball is

V = (4/3)pi* r^3

So..... the portion of the cylinder that is filled  by the ball (in terms of a ratio) is

[(4/3)pi * r^3] / [ 2 pi * r^3]  =  (4/3)/2 = 4/6 = 2/3 of the cylinder

So....the empty portion, in terms of ratio  = 1/3 of the cylinder 

23.  Using the Law of Cosines, we have

AP^2  = 15^2 + 12^2  - 2(15)(12)cos 24

AP = about 6.33

24. The diagonals of a rhombus bisect each other, therefore AP = PC  = 6.33

25. Since the diagonals bisect each other, BP = PD = 12. So, BD =24. And we can use the Law of Cosines to find AD

AD^2  = 24^2 + 15^2 - 2(24)(15)cos 24

AD  = about 11.97

So, using the Law of Sines

sin BDA / AB = sin 24 / AD

sinBDA / 15 = sin 24 / 11.97

sin^-1(15 sin 24 / 11.97) = BDA = 30.64°

26. We can use the Law of Cosines to find ACB....note BC = AD = 11.97  and AC = 2(AP) =2(6.33) = 12.66

AB^2 = BC^2 + AC^2 - 2 - 2(BC)(AC)cosACB

15^2 = 11.97^2 + 12.66^2 - 2(11.97)(12.66)cosACB

cos^-1 = (15^2 - 11.97^2 - 12.66^2) / (-2(11.97)(12.66)) = ACB = 74.98°

(–3/8 Log2(3/8)) + (– 5/8 Log2(5/8))=

(-3/8) [log₂ 3 - log₂ 8] + (-5/8) [log₂ 5 - log₂  8] =

(-3/8) [log₂ 3  - 3] + (-5/8) [ log₂ 5  - 3] =

-3 [ -3/8 -5/8]  - [ (3/8) log₂ 3 + (5/8)  log₂ 5 ] =

3  - [ 1/8)] * [ 3 log₂ 3 + 5 log₂ 5 ]  =

3 - [1/8] * [ log₂ 3^3 +  log₂ 5^5] =

3 - [1/8] * [ log₂ 27 + log₂ 3125] =

3 - [1/8] * [log₂ (27 * 3125)] =

3 - [log₂ 84375] / 8 =

[24 -  log₂ 84375 ] / 8 =  

 [about  0.954434]

This answer has been modified, 

Thanks CPhill and Alan for letting me know of my careless error :))

(–3/8 Log2(3/8)) + (– 5/8 Log2(5/8)) =

$$\\(-\frac{3}{8}Log_2(\frac{3}{8})) + (-\frac{5}{8} Log_2(\frac{5}{8})) \\\\
=-\frac{1}{8}[3Log_2(\frac{3}{8}) + 5 Log_2(\frac{5}{8})] \\\\
=-\frac{1}{8}[Log_2(\frac{3}{8})^3 + Log_2(\frac{5}{8})^5] \\\\
=-\frac{1}{8}[Log_2((\frac{3}{8})^3*(\frac{5}{8})^5 )] \\\\
=-\frac{1}{8}[Log_2(\frac{(3^3*5^5)}{8^8})] \\\\
=-\frac{1}{8}[Log_2 (3^3*5^5)-Log_2 8^8] \\\\
=-\frac{1}{8}[Log_2 (27*3125)-8Log_2 8] \\\\
=-\frac{1}{8}[Log_2 (84375)-8Log_2 2^3]\\\\
=-\frac{1}{8}[Log_2 (84375)-24Log_2 2]\\\\
=-\frac{1}{8}[Log_2 (84375)-24]\\\\
=3-\frac{[Log_2 (84375)]}{8}\\\\$$

This is the same as Chris's answer below.  :)

The question is finished but if you want an approximation ...

 this can be entered into the web 2 calc like this

3-log(84375,2)/8

$${\mathtt{3}}{\mathtt{\,-\,}}{\frac{{{log}}_{{\mathtt{2}}}{\left({\mathtt{84\,375}}\right)}}{{\mathtt{8}}}} = {\mathtt{0.954\: \!434\: \!002\: \!924\: \!964\: \!7}}$$

Most calcs only do base 10 or base e so this is how you would do it on one of those (uaing change of base rule)

3-(log(84375)/log(2))/8

$${\mathtt{3}}{\mathtt{\,-\,}}{\frac{\left({\frac{{log}_{10}\left({\mathtt{84\,375}}\right)}{{log}_{10}\left({\mathtt{2}}\right)}}\right)}{{\mathtt{8}}}} = {\mathtt{0.954\: \!434\: \!002\: \!924\: \!964\: \!9}}$$

Thanks Alan,

I almost missed this post - how did that happen? 

Thanks for finding this grapher.  It looks really neat.    

I'll add that 3D grapher to our reference material.  I'll add mine too.  I only just thought to do that.   

I didn't think of that Alan, yes there must be a distribution for which B is correct. 

C and D are more than 50% so they will never  be correct!

This assumes the distribution is a normal (Gaussian) one, though the question doesn't say so.  If it were a symmetrical triangular distribution with those parameters, for example, none of the displayed solutions would be right.

I wonder if there's a symmetrical distribution of some sort for which (b) would be the right answer!

.

just write 19 and youll pass....my guarantee