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As follows:

wrong

May I ask how you got that?

A simpler form of this fraction is $\dfrac{x + \sqrt{3}}{1 - x\sqrt{3}}$. Can you take it from here?

The answer is [200,360).

I checked at bartebly.com and they had the right answer

If we break everything into little triangles, (ABO, BOC, etc) we'll notice that ABMO = 1.5 triangles, and there are 8 triangles in total.

1.5/8 = 3/16

=^._.^=

Im sorry but that is incorrect

Refer to the graph of y = sin x or y = cos x to find the exact values of x in the interval [0, 4𝜋] that

satisfy the equation.
3 sin x = − 3/2

Hello Guest!

\(x\in \{\frac{7\pi}{6},\frac{11\pi}{6},\frac{19\pi}{6},\frac{23\pi}{6}\}\)

  !

Find the solutions of the equation that are in the interval [0, 2𝜋).

Hello Guest!

\(tan^2 x sin x = 3 sin x\\ sinx(tan^2x-3)=0\\ sinx=0\\ \color{blue}x\in \{\ 0,\pi ,2\pi\}\\ tan^2x-3=0\\ tan^2x=3\\ tan\ x=\pm \sqrt{3}\\ x=atan(\pm \sqrt{3})\)

\(\color{blue}x\in \{\frac{\pi}{3},\frac{2\pi}{3},\frac{4\pi}{3},\frac{5\pi}{3}\}\)

  !

\((2x+{3\over 2}(4x-6))-4(-(2-x)+{x\over 2})\)

                       \(=(2x+6x-9)-4(-2+x+{x\over 2})\)

                       \(=8x-9+8-6x\)

                        \(=2x-1\)

~Amy 

Suppose she has \($x\) at first.

Amt. of money she gives away = \(${2x\over 3}\)

⇒ Amt. of money she keeps = \(${x\over 3}\)

 ∴ She keeps \($50\)  in her left pocket and   \($({x\over 3}-50)\)  in her right pocket.                          [it is not really relevant] 

Now, if she gives away \($200\), 

Amt. of money she keeps = \($(x-200)\) 

According to question, 

\(x-200>{x\over 3}\)

⇒\({x\over 3}

⇒\(x<3x-600\)                            

⇒\(2x<600\)

⇒\(x<300\)

Also, Jeri has atleast \($200\) but less than \($300\)

∴ The possible values of x are (200,300). 

P.S. Corrections and suggestions are welcome. 

~Amy 

No, its absolutely right.

• Look, we arrived at the solution that Hugo had \(88\) with him.  
• He gave 1/4 of his money & an additional 1 to Calvin i.e. \(23\) 
• Now he has \(88-23=65\) remaining with him. 
• Then he gave 1/5 of remaining and an additional 4 to Kenneth i.e \(17\)
• He has \(65-17=48\) remaining with him
• Now its given that "amount of money Hugo had left in the end was $31 more than the amount of money he gave to Kenneth"
• So, Hugo gave \(17\) to Kenneth and he has \(48\) remaining with him. 

I did reverse answer check here and nothing came out wrong

See? everything fits into the question now what's wrong?

Okay, what's wrong?

There's a slight correction in the given GP, shouldn't it be \(2+r+{r^2\over 2}+{r^3\over 4}+...?\)

As per the question, 

First term of the GP   \(a=2\)

and common ratio  \(={r\over 2}\)

We know that sum of infinite  GP 

             S_{∞ \(={a\over r-1}\)}

⇒ S_∞  \(={2\over {r\over 2}-1}\)

            \(={4 \over r-2}\)

Now, S_∞ is defined for \(-1 < r <1 \)

∴ All possible values of r are in the interval  \((-1,1)\). 

~Amy 

That's wrong

Suppose Hugo had \(x\) money at first.

Hugo gave 1/4 of his money and an additional 1 to Cavin

Remaining money with Hugo \(=x-{1\over 4}x-1\)

                                               \(={3\over 4 }x-1\)

He gave 1/5 of the remaining money and an additional 4 to Kenneth

Remaining money with Hugo \(= ({3\over 4}x-1)-{1\over 5}({3\over 4}x-1)-4\) 

                                               \(={3\over 4}x-1-{3\over 20}x+{1\over 5}-4\)

                                               \(={12\over 20}x-{24\over 5}\)

Now according to question, 

⇒\({12 \over 20}x-{24\over 5}={1\over 5}({3\over4}x-1)+4+31\)

⇒\({12\over 20}x-{24\over 5}={3\over 20}x-{1\over 5}+35\)

⇒\({12\over 20}x-{24\over 5}={3\over 20}x+{174\over 5}\)

⇒\({9\over 20}x={198\over 5}\)

⇒\(x=88\)

∴ Hugo had 88 at first.

Now, amt. of money Hugo gave to Calvin \(={1\over 5}({3\over 4}x-1)+4\)

                                                                  \(={1\over 5}(65)+4\)

                                                                  \(=17\)

Thus, Hugo gave 17 units money to Calvin. 

~Amy                                         

Are you sure its an arithmetic sequence shouldn't it be a geometric sequence? 

In which case, 

sequence should be : \(3^2, x, 3^6\)

∴ Value of \(x= 3^4\)

First, find the common ratio: (2/3)/(3/2) = 4/9

Second, factor the infinite geometric series: (3/2)*((4/9)^0 + (4/9)^1 + (4/9)^2 + (4/9)^3 + ... )

Third, use the formula 1/(1-x) to solve for ((4/9)^0 + (4/9)^1 + (4/9)^2 + (4/9)^3 + ... ), where x = 4/9:  1/(1-4/9) = 9/5

Lastly, simplify: (3/2)*(9/5) = 27/10 or 2.7

The value of n is 14.

The distance XY is 12.

PLEASE tell me if this was helpful

and tell me how I can make my explanations better

:)