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Can somebody help me with this please? Heureka, Gingerale, Melody, anybody? Thanks.

I do not believe so; there is no reason not to simplify the fraction completely. 

As follows:

.

Thanks

I do not think our guest has interpreted the question as it is written, I think he/she has found the 1000th odd integer, not the 1000th integer with an odd number of digits.

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there are 9 one digit numbers.  1-9                                      9-1+1=9

there are  900   three digit numbers  100-999                          999-100+1=900

that is 909 altogether

So the 1000th one will have 5 digits.

1000-909=91

10000 is the 910th one, 

10090 is the 1000th one.

the prime factors of 2010 are 2,5,3 and 67

2010=2*5*3*67

In order for a fraction to be a terminating decimal you must first write it in its simplest from

THEN the prime factors of the denominator must be only 2 and 5.

so if   n/2010 is to terminate then the 3 and 67 must be cancelled out when the fraction is simplified.

3*67=201.

So the smallest positive interger value of n is 201 if the decimal is to terminate.

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you do not really need to find the prime factors of 2010

You can just factor out the powers of 10 then the powers of 2 and 5 and see what is left.

2010 = 10*201             

201 is not a multiple of 2 or 5 so 201 must be the smallest positive numerator if this fraction is going to terminate.

For some reason LaTex is not displaying properly.

If you cannot make sense of my answer let me know and I will try to write it without LaTex. :/

Hi MapleTheopry,

This is an unusual question. :)

\(Plog_{400}5+Qlog_{400}2=3\\ log_{400}(5^P)+log_{400}(2^Q)=3\\ log_{400}(5^P*2^Q)=3\\~\\ 400^3=5^P*2^Q\\ (2^4*5^2)^3=5^P*2^Q\\ 2^{12}*5^6=2^Q*5^P\\ so\\ P+Q=6+12=18 \)

Hahaha!!!

U 2.          Melody        

Your welcome

Also... the LaTeX code is   \leq  .  

Solve the following equations for  θ , in the the interval  0 < θ ≤  360° .

a)   sin θ + cos θ   =   0

                                                     Factor  cos θ  out of the terms on left side

(cos θ)( sin θ / cos θ + 1)   =   0

                                                     We can divide both sides by  cos θ  since  cos θ  can't be zero.

sin θ / cos θ  +  1   =   0

tan θ  +  1   =   0

tan θ   =   -1

And this occurs at     θ  =  135º     and     θ  =  315°

b)   3 cos θ   =   -2

                                    Divide both sides by  3 .

cos θ   =   -2/3

                                    Take the inverse cosine of both sides of the equation.

θ   =   arccos( -2/3 )

                                    By putting this into a calculator, I get

θ   ≈   131.81°

And we know that the cosine is also negative in the third quadrant, so..

θ   ≈   360° - 131.81°   ≈   228.19°     also has a cosine of  -2/3 .

c)   (sin θ - 1)(5 cos θ + 3)   =   0

Here, we need to set each factor equal to zero and solve each for  θ  .

First factor:

sin θ - 1   =   0

sin θ   =   1         There is only one angle that has a sin of 1 .

θ   =   90°

Second factor:

5 cos θ + 3   =   0

5 cos θ   =   -3

cos θ   =   -3/5

θ   ≈   126.87°          and          θ   ≈   360° - 126.87°   ≈   233.13°

d)   tan θ   =   tan θ ( 2 + 3 sin θ )

                                                             Subtract  tan θ  from both sides.

0   =   tan θ ( 2 + 3 sin θ )  -  tan θ

                                                             Factor  tan θ  out of the terms on the right side.

0  =  (tan θ)( (2 + 3 sin θ)  -  1 )

0  =  (tan θ)( 1 + 3 sin θ)

                                                      Set each factor equal to zero and solve for  θ .

tan θ   =   0    This occurs at..

θ   =   180°     and     θ   =   360°

1 + 3 sin θ   =   0

3 sin θ   =   -1

sin θ   =   -1/3

θ   ≈   -19.47° + 360°   ≈   340.53°          and          θ   ≈   180° - -19.47°   ≈   199.47°

If you're using a mac, you can do alt+< or > to get ≤ or ≥

Use the formula for solving the sum of degrees of the interior angles, then divide by \(n\) to find out only one of the interior angles.

Use this formula: \(\frac{(n-2)\times180}{n}\)

I'll do the first one: 

\(n=3\)

\(\frac{(3-2)*180}{2}=\frac{180}{2}=90\)

90 is an integer, so 3 will not work.

Keep on doing this until you get to 9.

You can use this online Math keyboard to write the Math symbols you want:

http://math.typeit.org/

a repeated root

does he mean like finding the value of k and then putting the answer as \(\frac{5^2}{8}\)

Rahuan, I am not completely sure what it means either, but I will interpret "double root" as a zero with a multiplicity of two. Assuming the above is true, we can use the property of the discriminant of a quadratic, \(b^2-4ac\) . The discriminant, if equal to zero, results in a quadratic with a "double root." 

or do u mean square root/ just (x²). (x = any number/ coefficients etc)

I understood until the equation part but what do u mean by double root?

201/2010=0.1