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Thank you!

\(\ \phantom{=\quad}\dfrac{10^{2000}+10^{2002}}{10^{2001}+10^{2001}}\\~\\ =\quad\dfrac{10^{2000}(1+10^{2})}{10^{2000}(10+10)}\\~\\ =\quad\dfrac{(1+10^{2})}{(10+10)}\\~\\ =\quad\dfrac{(1+100)}{20}\\~\\ =\quad\dfrac{1}{20}+\dfrac{100}{20}\\~\\ =\quad\dfrac{1}{20}+5 \)

The whole number closest to  5  and  1/20th  is  5 .

Not much! 

yeah

Awesome! Thanks for the help and the effort put in! However, I don't really understand how you got the values for height of person (b & 8cos theta) or why it works.

Thank you so much! :)

Hey, Manuel!!!

A concentric circle that has a radius greater than the given circle

P(A l B)  =       .24                                                  P( B l A)  = .24

                _____________  =   4/7  = .57                                ____  =   3/5  = .60

                       .42                                                                      .40

P(A)  = .40

P(B) = .42

Not independent because   

P(A l B)  not equal P(A)     and P(B l A)  not equal to B

They travel for 3 hours.....and the total distance covered = 240 miles

Let x be the rate of the slower car    and x + 10  be the rate of the faster

And    R*T  = D    so

3(x) + 3(x + 10) = 240

3x +  (3x + 30)  = 240

5^n + 6*7^n+ 1

Show that it's true for  n = 1

5^1 + 6*7^1 + 1  =

5 + 42 + 1  =  48     which is divisible by 8

Assume that it's true for  k ...that is

5^(k) + 6*7^k + 1     is divisible by 8

Prove that it is true for k + 1

5^(k + 1)  + 6*7^(k + 1) + 1

5*5^k + 7*6*7^k + 1

5*5^k + 42* 7^k + 1

5^k + 4*5^k + 6*7^k + 36*7^k + 1

(5^k + 6*7^k + 1)  +   4*5^k + 36*7^k

(5^k + 6*7^k + 1) +  4*(5^k + 9*7^k)

Note that  5^k  is odd   and so is 9*7^k  ....but the sum of two odds is even = 2n

So we have

(5^k + 6*7^k + 1) + 4 (2n)  =

(5^k + 6*7^k + 1)  + 8n

The first was assumed to be divisible by 8   and the second term is also divisible by 8

I think I like your method better, hectictar  ....!!!

15bx - 20  >  35

                                  We can factor  -5  out of  15bx - 20  because  -5( -3bx + 4 )  =  15bx - 20

-5( -3bx + 4 )  >  35

                                  Rearrange the terms inside the parenthesees...

-5( 4 - 3bx )  >  35

                                  Divide both sides of the inequality by  -5 , a negative number, so swap the sign.

4 - 3bx  <  -7

Does that help answer your question?

Let the x-coordinate  Q  =  a

Then the y-coordinate of  Q  =  -a + 6

slope of AQ  =  slope between  (10, -10)  and  (a, -a + 6)\(\ =\ \frac{(-a+6)-(-10)}{a-10}\ =\ \frac{16-a}{a-10}\)

slope of OQ  =  slope between  (0, 0)  and  (a, -a + 6)\(\ =\ \frac{(-a+6)-(0)}{a-0}\ =\ \frac{6-a}{a}\)

in order for  m∠OQA  to be  90°,  the slope of AQ  must be the negative reciprocal of the slope of  OQ . So...

\(\frac{16-a}{a-10}\ =\ -(\frac{a}{6-a})\\~\\ \frac{16-a}{a-10}\ =\ \frac{-a}{6-a}\\~\\ (16-a)(6-a)\ =\ (-a)(a-10)\\~\\ 96-22a+a^2\ =\ -a^2+10a\\~\\ 2a^2-32a+96\ =\ 0\\~\\ a^2-16a+48\ =\ 0\\~\\ (a-12)(a-4)\ =\ 0\\~\\ a=12\qquad\text{or}\qquad a=4\)

There are two points for  Q  that make  m∠OQA = 90° .   They are:   (12, -6)   and   (4, 2)

12 + -6  =  6     and     4 + 2  =  6

6 + 6  =  12

I think that this is supposed to be    \(\displaystyle 5^{n}+6.7^{n}+1,\)

so let

\(\displaystyle f(n) = 5^{n}+6.7^{n}+1,\)

then

\(\displaystyle f(n+1)=5^{n+1}+6.7^{n+1}+1=5.5^{n}+42.7^{n}+1,\)

in which case

\(\displaystyle f(n+1)-5f(n)=(5.5^{n}+42.7^{n}+1)-5(5^{n}+6.7^{n}+1) \\ \hspace{72pt}=12.7^{n}-4,\)

so

\(\displaystyle f(n+1)=5f(n)+4(3.7^{n}-1).\)

The second term on the rhs is a multiple of 8, 

(3.7^n is odd so 3.7^n - 1 is even, and that's multiplied by 4)

so if the first term on the rhs is also a multiple of 8 the term on the lhs will be a multiple of 8.

i.e. if f(n) is divisible by 8 then f(n + 1) will also be divisible by 8.

Since f(1) = 5 + 6.7 + 1 = 48 is divisible by 8 it follows that f(2), (= 5^2 + 6.7^2 + 1 = 25 + 6.49 + 1 = 320) will be divisible by 8 and so then will

f(3), (= 5^3 + 6.7^3 + 1 = 125 + 6.343 + 1 = 125 + 2058 + 1 = 2184), etc.. 

f(n) will be divisible by 8 for all positive integers.

Adios Amigo! Have fun!

I will! Bye!

Figure this again

Area of rectangles  = 112

Area of triangles  =  (1/2) (24 + 24 )

Key - Rect  !!!!

Let Q  = (x, -x + 6)

So....by the Pythagorean Theorem...

[Distance from (0,0) to Q]^2  + [ Distance from (10, -10) to Q ]^2  = [Distance from (0,0) to (10, -10)]^2

 [ (x )^2 + ( -x + 6)^2] + [ (10-x)^2 + (-10 - (-x + 6)*^2 ]  = 200

[ (x^2) + (-x + 6)^2] + [ (10 - x)^2 + ( x - 16)^2] = 200

x^2 + x^2 - 12x + 36 + x^2 - 20x + 100 + x^2 - 32x + 256 = 200

4x^2 - 64x + 192  = 0    divide through by 4

x^2 - 16x + 48 = 0    factor

(x - 12) ( x - 4)  = 0

Set each factor to 0, solve for x    and we get that  x = 12  or x = 4

When x = 4, y = 2

When x = 12, y = -6

So...Q  =  (4, 2)   or Q = ( 12, - 6)

So....the sum of the coordinates =  12

Here's a graph :

Ah yes! Thank you!

\(3\ =\ \frac{15\sqrt{24}}{\sqrt{x}}\)

                        Divide both sides of the equation by  15

\(\frac{3}{15}\ =\ \frac{\sqrt{24}}{\sqrt x}\)

                                        Square both sides of the equation.

\(\Big(\frac{3}{15}\Big)^2\ =\ \Big(\frac{\sqrt{24}}{\sqrt x}\Big)^2\)

                                        Simplify both sides using the rule  \(\big(\frac{a}{b}\big)^2\ =\ \big(\frac{a}{b}\big)\big(\frac{a}{b}\big)\ =\ \frac{a^2}{b^2}\)

.\(\frac{3^2}{15^2}\ =\ \frac{(\sqrt{24}\,)^2}{(\sqrt x\,)^2}\)

\(\frac{9}{225}\ =\ \frac{24}{x}\)

                              Now multiply both sides of the equation by  x

\(x\cdot\frac{9}{225}\ =\ 24\)

                                                  Multiply both sides of the equation by  \(\frac{225}{9}\)

\(x\cdot\frac{9}{225}\cdot\frac{225}{9}\ =\ 24\cdot\frac{225}{9}\)

\(x\ =\ 24\cdot\frac{225}{9}\)

\(x\ =\ 600\)

The second one??? 

129 would be the answer 

OK...you got the rectangles correct

Remember that the area of a triangle   = (1/2) base * height

Refigure the areas of the two triangles

Area of square   =   side^2

Can you spot the correct answer  ???