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I am starting to doubt that there are no solutions for the seven remaining

tan(R) = 5 * sin(R)

Angle R ≈ 78.46304º

PQ = sin(R) * QR

The answer is 3.

1. The area of the parallelogram is 3*sqrt(&).

2. The point is (3,-5,-1).

PQ = 8*sqrt(3).

You're welcome. :))

=^._.^=

Thank you for the reply!

Hello :))

In Q and R there are 9. 

In Q and S there are 4. 

In R and S there are 11. 

Combined, there are 18 elements, so there are 3 elements that are the same (9+4+11-18)/2. 

I find it easier to visualize it through a venn diagram. 

=^._.^=

Lol "no rush" and solved in 7 minutes

we use the distant formula to found out that the diameter is 4 then we divide it by 2 and found out that the radius is 2 the area of this circle is 4\( \pi\)

If \(5-i\sqrt[4]{3}\) is a root, then we can set \(5+i\sqrt[4]{3}\) to also be a root, since it is the conjugate of the other root and will multiply to cancel out the imaginary term. It would look something like this:

\((x-(5-i\sqrt[4]{3}))(x-(5+i\sqrt[4]{3}))\\=x^2-10x+25+\sqrt{3}\\=(x-5)^2+\sqrt{3}\)

Again, we could use the same conjugate trick as before to cancel out the radical:

\(((x-5)^2+\sqrt{3})((x-5)^2-\sqrt{3})\\=(x-5)^4-3\\=x^4 - 20 x^3 + 150 x^2 - 500 x + 625-3\\ =\boxed{x^4 - 20 x^3 + 150 x^2 - 500 x + 622}\)

(note that for the second last step you could use the binomial theorem to expand the terms faster.

Let's say that the first term is "a", and the common ratio is "x".

Then the sequence is: \(a,ax,ax^{2}, ax^{3},...\).

The sequence will go on forever, and given that they are positive integers, I do not think it is possible to find the sum.

However, we can find the first term. We know that the second term is 45 and the fifth term is 32805, so:

\(ax=45\) and \(ax^{4}=32,805\).

From this, we can find x=27 and a=5/3.

(You might need Google Translate to understand)

(Es posible que necesite el Traductor de Google para comprender)

5040 since when we expand it it is 5040

According to Google Translate, the problem reads (según Google Translate, el problema dice): 

The second term of a geometric progression is 45 and the fifth is 32,805 calculate the first term, construct the progression and calculate the sum of them.

1/ ( a + 1)   +  1 / ( b + 1)   =

     a + b  +  2

_______________

ab + a +   b +  1

By Vieta

The  sum of the roots    =  11 / 5   =  a + b

And  the product of the roots =   4/5    =  ab

So

(11/5)  + 2                               21 / 5                 21

_________________   =   __________  =     ____ 

 (4/5)  + ( 11/5)  + 1                20  / 5                20

The problem is (in LaTeX):

Let \(a\) and \(b\) be the solutions to \(5x^{2}-11x+4=0\). Find \(\frac{1}{a+1} + \frac{1}{b+1}\).

FYI: In LaTeX, 

\binom{a}{b} = \(\binom{a}{b}\)

a^{b} = \(a^{b}\)

\frac{a}{b} = \(\frac{a}{b}\)

actually, I think i got the answer, it is     \(72\pi-72\sqrt{3}\)

The  terms  are

m^6  +  6m^5n +  15m^4n^2   +  20n^3n^3  +  15m^2n^4  +  6mn^5  + n^6

The terms in red  are odd

we first use the quadratic equation to found out what is A and B then we plugin number and use a calculator to found the answer which is 21/20 sorry for the previous answer, I see the question wrong

No worries!

Don't know how to solve this one, MG.......maybe someone else does....(sorry  !!! )

thx for the help 

odd + odd is always equal to even when an even number rise to an even number is always even then the answer is none of the turns is odd. 

Sum of  first  n psitive integers =   n ( n + 1)  /  2

So

1000   (1001)  /  2  = 

500 * 1001

500500    !!!!

Correct !!!