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I would translate each of the percentages into a common fraction:

          33 1/3%  =  1/3

          66 2/3%  =  2/3

          11 1/9%  =  1/9

          50%        =  1/2

and then multiply the common fractions together:

          for example:  33 1/3%  x  66 2/3%  =  1/3 x 2/3  =  2/9

and then translate that common fraction back into a percentage:

          for example:  2/9  =  22 2/9%

5^1/5 x 25^1/25 x 125^1/125 x 625^1/625 x ...

=  5^1/5 x 5^2/25 x 5^3/125 x 5^4/625 x ...

=  5^{[ 1/5 + 2/25 + 3/125 + 4/625 + ...}

Looking at only the exponent:

   1/5 + 2/25 + 3/125 + 4/625 + ...

=  1/5 + 2/5² + 3/5³ + 4/5⁴ + ...

=  1/5 + (1/5² + 1/5²) + (1/5³ + 1/5³ + 1/5³) + (1/5⁴ + 1/5⁴ + 1/5⁴ + 1/5⁴) + ...

=  1/5 + 1/5²              + 1/5³                          + 1/5⁴                  + ...               = (1/5) / (1 - 1/5)  =  1/4

                      + 1/5²               + 1/5³                           + 1/5⁴      + ...               = (1/5² / (1 - 1/5²)  =  1/20

                                                          + 1/5³                         + 1/5⁴  + ...        = (1/5³ / (1 - 1/5³)  =  1/100

                ....

Summing:  1/4 + 1/20 + 1/100 + ...  =  (1/4) / (1 - 1/5)  =  5/16

So the answer is:  5^5/16     

I'm not sure what scaling means, but if it means what I think it means, consider the following.  

Take a square, with sides S.  The area is (S • S) = S².  

Scale it by a.  Now the sides are (S • a) and the area is (S • a) • (S • a) = S²a².  

Make it a rectangle, scale by a, and the areas is (L • a) • (W • a) = LWa²  

Try a circle of radius R, scale by a, and the area is π • (Ra)² = πR²a²  

Anything that is equal to, or bigger than 65.

Let X be a square with side length 3 and A be 4. Scaling by a, we have a 12 x 12 square now. The area of this square is 12 x 12 = 144, \(4^2\) times bigger than the original square's area with 9. Thus, scaling by a increases the area by \(a^2\). (Note: If you still don't believe me, try this out with a different shape and different values for X and A. 

AB = BC = CD by the definition of square ( I didn't forget DA ). 

Thus we can use distance formula to calculate length AB, which has length sqrt(34) units, and slope -5/3. Thus CD should also have slope -5/3, and length sqrt(34) units. 

Point D should be on line y = -5x/3 - 4. DA has equation y = 3x/5 + 2.8 (perpendicular).

Thus we can solve for the x coodinate of D, which is obtained by using transitive property:

Point D has coordinates (-3,1)

x = -3

y = 1

u could also use desmos lol 

Can you show a diagram or where the "shaded region" is? I would like a visual please.

[-2, -4/(3)] should be the answer

The difference of the means is not significant because the re-randomizations show that it is within the range of what could happen by chance.

1)  angle(R) + angle(P) + angle(Q)  =  180^o

     --->         angle(R) + 107^o + 42^o  =  180^o  

     --->                               angle(R)  =  31^o  

     p / sin(P)  =  r / sin(R)     --->     p / sin(107^o)  =  49 / sin(31^o)  

                                            --->     p  =  49 x sin(107^o) / sin(31^o)  

                                            --->     p  =  ......

The other two practice problems are solved in the same manner.

bye bye

And what if the inequality sign is flipped >

Does that mean all answers bigger than 144?

Given right triangle(ABC) with right angle(ABC) and altitude(BD).

Since AD = 4  and  BD = 2,  use the formula:  AD/BD  =  BD/DC

--->     4/2  =  2/DC

and finish.

Thanks so much! I'm going to ace this homework!

speed is distance divided by time:  speed  =  1254 km / 3 hr  =  418 km/hr

Can you post a diagram or a description of the particular problem?

Since  sqrt(144) = 12  but you want answers less than 12, you will need to choose all the

possibilities that are smaller than 144.

Multiply m^2 by 10^-4 to get hectares.

Like so:

Sequence:  \(\frac{1}{8}, \frac{1}{8}+d, \frac{1}{8}+2d, \frac{1}{8}+3d, \frac{1}{5}\)

Last term:  \(\frac{1}{8}+4d=\frac{1}{5}\)  Use this to find d, then you can find and sum the three middle fractions.

You can write  \(e^{i\theta}=\cos \theta+i\sin \theta\), so the real part of each number (a, c and f) is obtained using \(\cos \theta\) and the imaginary part (b, d and g) using \(\sin \theta\).

Hi Geno, That's a good one, and I did think of it.  I didn't submit it because,  

although the problem isn't explicit about not using the same number in both  

groups (i.e., the "1"), I imposed that limitation on myself. 

_.