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a and b are positive integers such that \(ab + 5a + b = 2000\).
Find b.

\(\begin{array}{|rcll|} \hline ab + 5a + b &=& 2000 \\ (a+1)(b+5)-5 &=& 2000 \\ \mathbf{(a+1)*(b+5)} &=& \mathbf{2005} \\ && \text{find all integer products }\ (a+1)*(b+5) = 2005 \\ \hline \end{array} \)

All integer products:

\(\begin{array}{|ccl|r|r|r|r|l|} \hline && (a+1)*(b+5) & (a+1) & (b+5) & a & b & \text{$a$ and $b$ are positive integers} \\ \hline 2005 &=& 1\times 2005 & 1 & 2005 & 0 & 2000 \\ 2005 &=& 5\times 401 & 5 & 401 & \mathbf{4} & \mathbf{396} & \checkmark \\ 2005 &=& 401\times 5 & 401 & 5 & 400 & 0 \\ 2005 &=& 2005\times 1 & 2005 & 1 & 2004 & -4 \\ \hline \end{array} \)

\(\text{$b+5 = 401$, so $b=401-5$ hence $\mathbf{b = 396}$ }\)

"A right triangle has an area of 24 and a perimeter of 40.  What is the hypotenuse of the right triangle?"

The sum of the first n terms of a sequence is \(n^2 + 5n\).  
What is the \(1000th\) term of the sequence?

Formula:  \(\mathbf{\boxed{s_n =\dfrac{(a_1+a_n)}{2}\cdot n }}\)

\(\begin{array}{|rcll|} \hline \mathbf{a_1} &=& \mathbf{s_1} \quad | \quad s_n =n^2 + 5n \\ &=& 1^2 + 5*1 \\ \mathbf{a_1} &=& \mathbf{6} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline s_n &=& \dfrac{(a_1+a_n)}{2}\cdot n \quad | \quad a_1=6,\ s_n = n^2 + 5n \\\\ n^2 + 5n &=& \dfrac{(6+a_n)}{2}\cdot n \\\\ 2n^2 + 10n &=& (6+a_n)n \quad | \quad :n \\ 2n + 10 &=& 6+a_n \\ 2n + 4 &=& a_n \\ \mathbf{a_n} &=& \mathbf{2n+4} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{a_n} &=& \mathbf{2n+4} \quad | \quad n=1000 \\ a_{1000} &=& 2*1000+4 \\ \mathbf{a_{1000}}&=& \mathbf{2004} \\ \hline \end{array}\)

See yesterday's answer here:  https://web2.0calc.com/questions/what-is-the-vector-that-results-from-transforming_1#r1 

Please help with this number theory problem: Find the largest integer n such that \(2^n \)divides \(17^9 - 9^9\).

My attempt:

Formula:

\(\begin{array}{|rcll|} \hline a^9-b^9 &=& (a^3-b^3)(a^6+a^3b^3+b^6) \quad | \quad (a^3-b^3) = (a-b)(a^2+ab+b^2) \\ \mathbf{a^9-b^9} &=& \mathbf{(a-b)(a^2+ab+b^2)(a^6+a^3b^3+b^6)} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{a^9-b^9} &=& \mathbf{(a-b)(a^2+ab+b^2)(a^6+a^3b^3+b^6)} \\\\ 17^9-9^9 &=& (17-9)(17^2+17*9+9^2)(17^6+17^39^3+9^6) \\ 17^9-9^9 &=& 8*523*28250587 \\ && \boxed{523 ~\text{ is odd and }~ 28250587~\text{ is odd}\\ \text{and } 8=2^3 } \\ 17^9-9^9 &=& 2^3*523*28250587 \\ \hline \end{array} \)

The largest integer n is 3

MN = MP = 1

Angle NMO is 30°

OM = MN / cos(30°)  = 1.154700538

OP = OM + MP = 2.154700538

The radius of a large circle    r = 2.154700538  

Three circles of radius 1 are drawn, each tangent to each other.  Find the radius of the large circle.

Find:

\(x = 1 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \ddots}}}}.\)

\(\begin{array}{|rcll|} \hline \mathbf{x} &=& \mathbf{1 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \ddots}}}}} \\ x-1 &=& \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \ddots}}}} \\ x-1 &=& \cfrac{1}{2 + x-1} \\ (x-1)(2 + x-1) &=& 1 \\ x+x^2-x-2-x+1 &=& 1 \\ x^2-2 &=& 0 \\ x^2 &=& 2 \\ \mathbf{ x} &=& \mathbf{\sqrt{2}} \\ \hline \end{array}\)

Just a random maths question...

I was bored and wanted help understanding a future topic.

Sorry mate but I don't know how to help.

One thing you could do is:

- Draw an equilateral triangle in the middle (each point connecting to the centre of the small circles)

--> The length of the triangle sides would = 2

- Use this formula to find the distance from the vertices of the triangle to the centre:

--> https://www.quora.com/What-is-the-distance-between-the-centre-of-an-equilateral-triangle-and-any-of-its-vertices 

- + 1 (length of the radius)

Hope this helps (pretty sure this is not the way to go about it but it logically works)..

:) <3

[R1] - [R2] - [R3] + [R4] = 140.

a - b = 32 pi - 6 pi = 26 pi.

Call the height of the trapezoid "x"   --->   AD  =  x

Since CD = 4, triangle(CDA) is a right triangle with CA  =  sqrt(4² + x²)  =  sqrt(16 + x²)

Since AB = 9, triangle(BAD) is a right triangle with BD  = sqrt(9² + x²)  =  sqrt(81 + x²)

The area of the trapezoid will be the area of triangle(CDA) + area triangle(CAB).

Let CA be the base of each of these triangles.

Since the diagonals are perpendicular, the sum of the heights of these triangles is BD.

Area of trapezoid(ABCD)  =  Area of triangle(CDA) + area triangle(CAB)  =  ½·sqrt(16 + x²)·sqrt(81 + x²).

Also, Area of trapezoid(ABC)  =  ½·x·(4 + 9)  =  (13/2)·x

Setting these two area equal to each other:  (13/2)·x  =  ½·sqrt(16 + x²)·sqrt(81 + x²).

--->      13x  =  sqrt(16 + x²)·sqrt(81 + x²).

--->   169x²  =  (16 + x²)·(81 + x²).

--->   169x²  =  1296 + 97x² + x⁴

--->           0  =  x⁴ - 72x² + 1296

--->           0  =  (x² - 36)²

--->           0  =  x² - 26

--->           x  =  6

Area  =  (13/2)·x  =  (13/2)·6  =  39

Since this is an isosceles trapezoid, angle(A) = angle(B)  and  angle(C) = angle(D).

So, the total angles:  (4x + 20) + (4x + 20) + (3x - 15) + (3x - 15)  =  360.

--->     14x + 10  =  360     --->     x  =  25.

Can you take it from here?

The sum of the yellow regions must be the same as the sum of the blue regions for this reason:

The diameter of each of the smaller circles is one-half the diameter of the larger circle.

To make it easy, call the radius of each of the smaller circle "1" so that the radius of the larger circle is "2".

The area of the larger circle is:  Area  =  pi · 2²  =  4·pi.

The area of each of the smaller circles is:  Area  =  pi · 1²  =  1·pi..

So, the total area of all the smaller circles (added together) is:  Area  =  4·pi.

Therefore, the total area of all the smaller circles is the same as the area of the larger circle.

So, the overlap of the smaller circles (the yellow areas) must equal the "underlap"  -- the area of the larger circle that isn't covered by the smaller circles -- the blue areas.

There are 4*P(5,2) = 240 ways to place an economics text book in the middle as well as place a biology textbook at each end.  There are 5! = 120 ways to arrange the remaining textbooks.

So the answer is 240*120 = 28800.

Nevermind found i!. It's \(\sqrt{2}\)

Let the crosspoint of the diagonals of the rhombus be an N.

Let the crosspoint of AD and BF be an M

BN = [sqrt(2² + 3²)] / 2 = 1.802775638

tan(∠FBD) = 2 / 3             ∠FBD = 33.69006753°

NM = BN * tan(∠FBD)   =   1.201850425

Shaded area       A= ( BN * NM ) * 2  =  4.3333333333 units squared  

Angles in a triangle add up to 180 

Thus, \(A+B+C=180\)

\((p+2q)+(6p-5q)+(5p+3q)=180\) 

\(12p=180\)

\(p=15\)

here's a hint: reflect the triangle around C so that AC is on BC and look for similar triangles

i'll write out a whole solution in a bit but that's mainly what you need to figure it out

If  a  =  (1, -3)  then  5a  =  (5, -15).

If  b  =  (-1, 2)  then  4b  =  (-4, 8).

4b - 5a  =  (-4, 8) - (5, -15)  =  (-9, 23).

Each side of each tube is a rectangle 4 ft x 11 ft  =  44 ft².

Since both the inside and outside are to be painted, each side  =  88 ft².

Since each tube is hexagonal:  88 ft² x 6  =  528 ft².

Since there are 12 tubes:  528 ft² x 12  =  6336 ft².

Since each can of point covers 40 ft²:  6366 ft² / 40 ft²  =  158.4 cans   --->   159 cans.

If there is a 40% chance that it will rain on Monday, then there is a 60% chance that it won't rain on Monday.

If there is a 30% chance that it will rain on Tuesday, then there is a 70% chance that it won't rain on Tuesday.

If there is a 50% chance that it will rain on Wednesday, then there is a 50% chance that it won't rain on Wednesday.

To rain on exactly two days:

-- rain on Monday (40%) and on Tuesday (30%) but not on Wednesday (50%)  =  0.40 x 0.30 x 0.50  =  0.06

-- rain on Monday (40%) and on Wednesday (50%) but not on Tuesday (70%)  =  0.40 x 0.50 x 0.70  =  0.14

-- rain on Tuesday (30%) and on Wednesday (50%) but not on Monday (60%)  =  0.30 x 0.50 x 0.60  =  0.09

Adding these individual situations:  0.06 + 0.14 + 0.09  =  0.29

You can draw a venn diagram to help....

Assume there is 100 kids   (100%)

75 + 30 + 15 = 120     there is 20 too many      20 % must ride both