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x   must  be  ≥  0

And   5 - sqrt x     must  be ≥  0

So

5 - sqrt x  ≥  0

5 ≥  sqrt x        square  both sides

25 ≥  x

So.....the  domain  is

[ 0 ,  25  ] 

See  the graph here  :  https://www.desmos.com/calculator/gxk17ddz6u

sin  5°   +   sin 10°  +  sin 15°  +   ..........  + sin 360°

Note  that  we  can write

(sin 5° + sin 355° )  + ( sin 10 °  + sin 350°)  + ( sin 15° +  sin 345°)  +  ....... + ( sin 175° + sin 185°) +.

(sin 180° )  + (sin 360°)   =

0   +   0    +   0  +  .........+  0  +  0   +    0    =   

0

Easy

x = 5y - 13 and x + y = 140.

Using substitution, 5y - 13 + y = 140 ====> 6y = 153, so y = 25.5, and x = 114.5

Help??

yup, check his stats

y - 1 = 2(x - 1) ==> y = 2x - 2 + 1 = 2x -1

Test the point (0, 0) ===> The equation should be y > 2x - 1

Oh really? I must have just been missing them the past few days. 

they still come online pretty often :D

As follows:

ProblemSolver is right, that would be mean, not median. Also, it is suspicious this is the exact same resposne (down to the same spelling mistakes) as the response given to the exact same question posted earlier by AnswerMachine:

https://web2.0calc.com/questions/help-plz-hard-problem

For positive real numbers and the equation
\(\arctan x + \arccos \frac{y}{\sqrt{1 + y^2}} = \arcsin \frac{3}{\sqrt{10}}\)
reduces to an equation of the form \(xy + ax + by + c = 0\).

My attempt:

\(\small{ \begin{array}{|rcll|} \hline \arctan x + \arccos \frac{y}{\sqrt{1 + y^2}} &=& \arcsin \frac{3}{\sqrt{10}} \quad | \quad \tan()~ \text{both sides} \\\\ \tan \Big(\arctan x + \arccos \frac{y}{\sqrt{1 + y^2}} \Big) &=& \tan \Big( \arcsin \frac{3}{\sqrt{10}} \Big) \\\\ \dfrac{ \sin \Big(\arctan x + \arccos \frac{y}{\sqrt{1 + y^2}} \Big) } { \cos \Big(\arctan x + \arccos \frac{y}{\sqrt{1 + y^2}} \Big) } &=& \dfrac{ \sin \Big( \arcsin \frac{3}{\sqrt{10}} \Big) } { \cos \Big( \arcsin \frac{3}{\sqrt{10}} \Big) } \\\\ \hline \dfrac{ \sin (\arctan x) \cos( \arccos \frac{y}{\sqrt{1 + y^2}} ) + \cos (\arctan x) \sin( \arccos \frac{y}{\sqrt{1 + y^2}} )} { \cos (\arctan x) \cos( \arccos \frac{y}{\sqrt{1 + y^2}} ) - \sin (\arctan x) \sin( \arccos \frac{y}{\sqrt{1 + y^2}} )} &=& \dfrac{ \sin \Big( \arcsin \frac{3}{\sqrt{10}} \Big) } { \cos \Big( \arcsin \frac{3}{\sqrt{10}} \Big) } \\\\ \hline \dfrac{ \sin (\arctan x) \frac{y}{\sqrt{1 + y^2}} + \cos (\arctan x) \sin( \arccos \frac{y}{\sqrt{1 + y^2}} )} { \cos (\arctan x) \frac{y}{\sqrt{1 + y^2}} - \sin (\arctan x) \sin( \arccos \frac{y}{\sqrt{1 + y^2}} )} &=& \dfrac{ \frac{3}{\sqrt{10}} } { \cos \Big( \arcsin \frac{3}{\sqrt{10}} \Big) } \\\\ \boxed{\sin (\arctan x) = \frac{x}{\sqrt{1 + x^2}} \\ \cos (\arctan x) = \frac{1}{\sqrt{1 + x^2}} } \\ \dfrac{ \frac{x}{\sqrt{1 + x^2}} \frac{y}{\sqrt{1 + y^2}} + \frac{1}{\sqrt{1 + x^2}} \sin( \arccos \frac{y}{\sqrt{1 + y^2}} )} { \frac{1}{\sqrt{1 + x^2}} \frac{y}{\sqrt{1 + y^2}} - \frac{x}{\sqrt{1 + x^2}} \sin( \arccos \frac{y}{\sqrt{1 + y^2}} )} &=& \dfrac{ \frac{3}{\sqrt{10}} } { \cos \Big( \arcsin \frac{3}{\sqrt{10}} \Big) } \\\\ \boxed{\sin( \arccos z ) = \sqrt{1 - z^2} \\ \cos( \arcsin z ) = \sqrt{1 - z^2} } \\ \dfrac{ \frac{x}{\sqrt{1 + x^2}} \frac{y}{\sqrt{1 + y^2}} + \frac{1}{\sqrt{1 + x^2}} \sqrt{1 - \frac{y^2}{1 + y^2} } } { \frac{1}{\sqrt{1 + x^2}} \frac{y}{\sqrt{1 + y^2}} - \frac{x}{\sqrt{1 + x^2}} \sqrt{1 - \frac{y^2}{1 + y^2} } } &=& \dfrac{ \frac{3}{\sqrt{10}} } { \sqrt{1- \frac{9}{10} } } \\\\ \hline \dfrac{ \frac{x}{\sqrt{1 + x^2}} \frac{y}{\sqrt{1 + y^2}} + \frac{1}{\sqrt{1 + x^2}} \frac{1}{\sqrt{1 + y^2}} } { \frac{1}{\sqrt{1 + x^2}} \frac{y}{\sqrt{1 + y^2}} - \frac{x}{\sqrt{1 + x^2}} \frac{1}{\sqrt{1 + y^2}} } &=& \dfrac{ \frac{3}{\sqrt{10}} } { \frac{1}{\sqrt{10}} } \\\\ \hline \dfrac{xy+1}{y-x} &=& 3 \\\\ xy+1 &=& 3(y-x) \\\\ xy-3(y-x)+1 &=& 0 \\\\ \mathbf{xy+3x-3y + 1} &=& \mathbf{0} \\ \hline a&=& 3 \\ b&=& -3 \\ c &=& 1 \\ \hline \end{array} }\)

Very impressive guest.

Let's see if I can find a quicker way...

1,2,2,3,3,4

A standard six-sided die is rolled 6 times.

You are told that among the rolls, there was one 1, 2 two's, two three 's, and one four.

How many possible sequences of rolls could there have been?

6!/(2!2!) = (2*3*4*5*6)/4 = 2*3*5*6 = 180 ways

or

How many ways to place the 2's     6C2 = 15

for each of those there are 4C2 ways to place the 3's = 6

then 2 ways to place the 4

and then only 1 way to place the 1

15*6*2=180 ways

So we have done it 3 ways in total and each time we got the answer 180 ways    

permutations of {1, 2, 2, 3, 3, 4}==6!/(2!)(2!)==180 such sequences as follows:

{{1, 2, 2, 3, 3, 4}, {1, 2, 2, 3, 4, 3}, {1, 2, 2, 4, 3, 3}, {1, 2, 3, 2, 3, 4}, {1, 2, 3, 2, 4, 3}, {1, 2, 3, 3, 2, 4}, {1, 2, 3, 3, 4, 2}, {1, 2, 3, 4, 2, 3}, {1, 2, 3, 4, 3, 2}, {1, 2, 4, 2, 3, 3}, {1, 2, 4, 3, 2, 3}, {1, 2, 4, 3, 3, 2}, {1, 3, 2, 2, 3, 4}, {1, 3, 2, 2, 4, 3}, {1, 3, 2, 3, 2, 4}, {1, 3, 2, 3, 4, 2}, {1, 3, 2, 4, 2, 3}, {1, 3, 2, 4, 3, 2}, {1, 3, 3, 2, 2, 4}, {1, 3, 3, 2, 4, 2}, {1, 3, 3, 4, 2, 2}, {1, 3, 4, 2, 2, 3}, {1, 3, 4, 2, 3, 2}, {1, 3, 4, 3, 2, 2}, {1, 4, 2, 2, 3, 3}, {1, 4, 2, 3, 2, 3}, {1, 4, 2, 3, 3, 2}, {1, 4, 3, 2, 2, 3}, {1, 4, 3, 2, 3, 2}, {1, 4, 3, 3, 2, 2}, {2, 1, 2, 3, 3, 4}, {2, 1, 2, 3, 4, 3}, {2, 1, 2, 4, 3, 3}, {2, 1, 3, 2, 3, 4}, {2, 1, 3, 2, 4, 3}, {2, 1, 3, 3, 2, 4}, {2, 1, 3, 3, 4, 2}, {2, 1, 3, 4, 2, 3}, {2, 1, 3, 4, 3, 2}, {2, 1, 4, 2, 3, 3}, {2, 1, 4, 3, 2, 3}, {2, 1, 4, 3, 3, 2}, {2, 2, 1, 3, 3, 4}, {2, 2, 1, 3, 4, 3}, {2, 2, 1, 4, 3, 3}, {2, 2, 3, 1, 3, 4}, {2, 2, 3, 1, 4, 3}, {2, 2, 3, 3, 1, 4}, {2, 2, 3, 3, 4, 1}, {2, 2, 3, 4, 1, 3}, {2, 2, 3, 4, 3, 1}, {2, 2, 4, 1, 3, 3}, {2, 2, 4, 3, 1, 3}, {2, 2, 4, 3, 3, 1}, {2, 3, 1, 2, 3, 4}, {2, 3, 1, 2, 4, 3}, {2, 3, 1, 3, 2, 4}, {2, 3, 1, 3, 4, 2}, {2, 3, 1, 4, 2, 3}, {2, 3, 1, 4, 3, 2}, {2, 3, 2, 1, 3, 4}, {2, 3, 2, 1, 4, 3}, {2, 3, 2, 3, 1, 4}, {2, 3, 2, 3, 4, 1}, {2, 3, 2, 4, 1, 3}, {2, 3, 2, 4, 3, 1}, {2, 3, 3, 1, 2, 4}, {2, 3, 3, 1, 4, 2}, {2, 3, 3, 2, 1, 4}, {2, 3, 3, 2, 4, 1}, {2, 3, 3, 4, 1, 2}, {2, 3, 3, 4, 2, 1}, {2, 3, 4, 1, 2, 3}, {2, 3, 4, 1, 3, 2}, {2, 3, 4, 2, 1, 3}, {2, 3, 4, 2, 3, 1}, {2, 3, 4, 3, 1, 2}, {2, 3, 4, 3, 2, 1}, {2, 4, 1, 2, 3, 3}, {2, 4, 1, 3, 2, 3}, {2, 4, 1, 3, 3, 2}, {2, 4, 2, 1, 3, 3}, {2, 4, 2, 3, 1, 3}, {2, 4, 2, 3, 3, 1}, {2, 4, 3, 1, 2, 3}, {2, 4, 3, 1, 3, 2}, {2, 4, 3, 2, 1, 3}, {2, 4, 3, 2, 3, 1}, {2, 4, 3, 3, 1, 2}, {2, 4, 3, 3, 2, 1}, {3, 1, 2, 2, 3, 4}, {3, 1, 2, 2, 4, 3}, {3, 1, 2, 3, 2, 4}, {3, 1, 2, 3, 4, 2}, {3, 1, 2, 4, 2, 3}, {3, 1, 2, 4, 3, 2}, {3, 1, 3, 2, 2, 4}, {3, 1, 3, 2, 4, 2}, {3, 1, 3, 4, 2, 2}, {3, 1, 4, 2, 2, 3}, {3, 1, 4, 2, 3, 2}, {3, 1, 4, 3, 2, 2}, {3, 2, 1, 2, 3, 4}, {3, 2, 1, 2, 4, 3}, {3, 2, 1, 3, 2, 4}, {3, 2, 1, 3, 4, 2}, {3, 2, 1, 4, 2, 3}, {3, 2, 1, 4, 3, 2}, {3, 2, 2, 1, 3, 4}, {3, 2, 2, 1, 4, 3}, {3, 2, 2, 3, 1, 4}, {3, 2, 2, 3, 4, 1}, {3, 2, 2, 4, 1, 3}, {3, 2, 2, 4, 3, 1}, {3, 2, 3, 1, 2, 4}, {3, 2, 3, 1, 4, 2}, {3, 2, 3, 2, 1, 4}, {3, 2, 3, 2, 4, 1}, {3, 2, 3, 4, 1, 2}, {3, 2, 3, 4, 2, 1}, {3, 2, 4, 1, 2, 3}, {3, 2, 4, 1, 3, 2}, {3, 2, 4, 2, 1, 3}, {3, 2, 4, 2, 3, 1}, {3, 2, 4, 3, 1, 2}, {3, 2, 4, 3, 2, 1}, {3, 3, 1, 2, 2, 4}, {3, 3, 1, 2, 4, 2}, {3, 3, 1, 4, 2, 2}, {3, 3, 2, 1, 2, 4}, {3, 3, 2, 1, 4, 2}, {3, 3, 2, 2, 1, 4}, {3, 3, 2, 2, 4, 1}, {3, 3, 2, 4, 1, 2}, {3, 3, 2, 4, 2, 1}, {3, 3, 4, 1, 2, 2}, {3, 3, 4, 2, 1, 2}, {3, 3, 4, 2, 2, 1}, {3, 4, 1, 2, 2, 3}, {3, 4, 1, 2, 3, 2}, {3, 4, 1, 3, 2, 2}, {3, 4, 2, 1, 2, 3}, {3, 4, 2, 1, 3, 2}, {3, 4, 2, 2, 1, 3}, {3, 4, 2, 2, 3, 1}, {3, 4, 2, 3, 1, 2}, {3, 4, 2, 3, 2, 1}, {3, 4, 3, 1, 2, 2}, {3, 4, 3, 2, 1, 2}, {3, 4, 3, 2, 2, 1}, {4, 1, 2, 2, 3, 3}, {4, 1, 2, 3, 2, 3}, {4, 1, 2, 3, 3, 2}, {4, 1, 3, 2, 2, 3}, {4, 1, 3, 2, 3, 2}, {4, 1, 3, 3, 2, 2}, {4, 2, 1, 2, 3, 3}, {4, 2, 1, 3, 2, 3}, {4, 2, 1, 3, 3, 2}, {4, 2, 2, 1, 3, 3}, {4, 2, 2, 3, 1, 3}, {4, 2, 2, 3, 3, 1}, {4, 2, 3, 1, 2, 3}, {4, 2, 3, 1, 3, 2}, {4, 2, 3, 2, 1, 3}, {4, 2, 3, 2, 3, 1}, {4, 2, 3, 3, 1, 2}, {4, 2, 3, 3, 2, 1}, {4, 3, 1, 2, 2, 3}, {4, 3, 1, 2, 3, 2}, {4, 3, 1, 3, 2, 2}, {4, 3, 2, 1, 2, 3}, {4, 3, 2, 1, 3, 2}, {4, 3, 2, 2, 1, 3}, {4, 3, 2, 2, 3, 1}, {4, 3, 2, 3, 1, 2}, {4, 3, 2, 3, 2, 1}, {4, 3, 3, 1, 2, 2}, {4, 3, 3, 2, 1, 2}, {4, 3, 3, 2, 2, 1}}==180 permutations.

301 , 303 , 305 , 307 , 309 , 311 , 313 , 315 , 317 , 319 , 321 , 323 , 325 , 327 , 329 , 331 , 333 , 335 , 337 , 339 , 341 , 343 , 345 , 347 , 349 , 351 , 353 , 355 , 357 , 359 , 361 , 363 , 365 , 367 , 369 , 371 , 373 , 375 , 377 , 379 , 381 , 383 , 385 , 387 , 389 , 391 , 393 , 395 , 397 , 399 , >>Number of terms = 50>>Total Sum == 17500

Hi xy   

Thanks for answering HSDx and guest     But, guest, my answer is for you to learn from as well    

What real value of t produces the smallest value of the quadratic t^2-9t-36?

maybe it will make more sense to you change the t to an x and then put it equal to y

\(y=x^2-9x-36 \)

Now you are being asked for the smallest value of x

     If you graph this you will get a concave up parabola.

    I know this becasue the x^2 makes it a parabola

    and the invisable number in front of the x^2 is  +1. 

    Since it is + the parabola will be concave up.

The lowest point of a concave up parabola will be the turning point. The x value for it which will be halfway between the roots.

roots are when y=0 so

\(0=x^2-9x-36\\ 0=(x-12)(x+3)\\ x=12\;\;or \;\;x=-3 \)

Halfway between the roots is  x= (12+-3)/2 = 4.5

That is the x value (actually the t value) that you want.     t=4.5

that is the value that will give the expression the smallest value.

If you wanted to find that smallest value then you would sub 4=4.5 back into the expression.

Hi,

I think that is the mean, not median 

Principle of inclusion exclusion

X/2 + x/3 + x/6 = 180

3x + 2x +x =180 * 6

6x = 180*6

x = 180
 

Sent from my Android with TapaTalk

AC = 10(sin30) + 10(cos30)

BC = 10(sin45)        or      10(cos45) 

4c3 = 4c1= 4

-mps101

x = 180 degrees

x/2 = 90º

x/3 = 60º

x/6 = 30º