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Thank you !

Starting position =  - 5/3/5 ft

Finishing position = 987/10

Difference =  987/10 - (- 5 3/5) ft  ==== you can do the rest

The center will be 1/2 the x dimension of 10 = 5

    and 1/2 the  y dimension  = 20

Radius = 60

h, k = 5,20

(x-5)^2 +(y-20)^2 = 60^2

You could also see that 1 = 13/15 + 10/75

However, CU and Guest's method would probably be better.

You are very welcome!

:P

Not sure what you're meaning....

Hi! Welcome to web2.0calc, if you're new!

I'm not sure if I understand this problem...

Here's all you reaaly need to know here

If we divide the power on i   by  4  and we get a remainder of 1,  the result is  i

If we divide the power on i by 4  and get  a remainder of 2, the result is  -1

If we divide the power on i by 4  and get a remainder of 3, the result is  - i

If we divide the power on i  by 4  and get a whole number, the result is  1

So

i^4  = 4/4  =  1    = a whole number....so....the result is 1

i^5   =  5/4  = 1 + 1/4   = remainder of 1, so the result is i

i^6  = 6/4  = 1 + 2/4  = remainder of 2, so the result is -1

i^7 = 7/4  = 1 + 3/4 =  remainder of 3, so the result  = -i

i^8  = 8/4  = 2....whole number ....result is  1

i^9 = 9/4  = 2 + 1/4  = remainder = 1  so the result = i

i^10 = 10/4  = 2 + 2/4 =  remainder of 2  ....so the result is -1

b. The pattern is, for any integer n ≥ 1

i^(4n-3)  = i

i^(4n - 2)  = -1

i^(4n - 1)  = -i

i^(4n)  = 1

c.  i^2001  =  2001/4   =  500 + 1/4  ....remainder  = 1....so....the result  =   i

Thanks Ginger, I've never been considered more numerate than a cat before. That is high praise.  

Simplify (1+i)^2016-(1-i)^2016

\(1+i=\sqrt2e^{\frac{\pi}{4}i} \qquad \text{When presented in polar form}\\~\\ (1+i)^{2016}=\left( \sqrt2e^{\frac{\pi}{4}i} \right)^{2016}\\ (1+i)^{2016}= (\sqrt2)^{2016}*(e^{\pi i} )^{504}\\ (1+i)^{2016}= (\sqrt2)^{2016}*(e^{2\pi i} )^{252}\\ (1+i)^{2016}= (\sqrt2)^{2016}*(e^{0 i} )^{252}\\ (1+i)^{2016}= (\sqrt2)^{2016}*(1 )^{252}\\ (1+i)^{2016}= (\sqrt2)^{2016}*1\\~\\ (1-i)^{2016}=\left( \sqrt2e^{\frac{-\pi}{4}i} \right)^{2016}\\ (1-i)^{2016}=(\sqrt2)^{2016} (e^{\frac{-\pi}{4}i})^{2016}\\ (1-i)^{2016}=(\sqrt2)^{2016} (e^{-2\pi i})^{252}\\ (1-i)^{2016}=(\sqrt2)^{2016} (e^{0 i})^{252}\\ (1-i)^{2016}=(\sqrt2)^{2016} *1\\ so\\ (1+i)^{2016}-(1-i)^{2016}=(\sqrt2)^{2016}-(\sqrt2)^{2016}=0 \)

sqrt (x) < 7      square both sides

x < 49

But  x also has to be  ≥ 0

So.....the solutions are on this interval

0  ≤  x  < 49

Square both sides of the equation and you will get:

x < 49      Does that help ya with the answer(s)?

(-8, -7) to (9,−6) that partitions the segment into a ratio of 3 to 1?

We can find the partition point thusly :

[ (-8) + (3/4)(9 - -8) ,  -7 + (3/4)(-6 - -7) ]  =

[ -8 + (3/4) (17) , -7 + (3/4)(1) ]  =

[-8 + 51/4 , -7 + 3/4] =

(19/4, -25/4)

\(\frac{n}{5}=k+\frac{1}{5}\qquad \text{k is an integer}\\ n=5k+1\\ 3n=15k+3\\ \frac{3n}{5}=3k\quad Remainder\;3 \)

screenshot added. :)

The question does not make sense to me.  Are you sure it is copied properly?

Can any one else throw any light on the meaning?

I have been asked to discuss this question further.

I have only looked at part A again. I assume the problem with part A is the same as that of part B.

Lizzie has a strange deck of 24 number cards that contains cards labeled 2 through 9 in three different colors (red, blue, and green).
(a) In how many ways can she draw one card from each color so that the sum of all the cards is 10?

The only triads that add to 10 are 

6,2,2

5,2,3

4,4,2

4,3,3

If you think there is a problem with this so far then let me know.

Consider 6,2,2

The 6 can be red blue or green  3 choices. Once the 6 is allocated a colour, both the other colours must be 2. 

               SO there are 3 ways to allocate 6,2 and 2 if each one must be a different colour.

The same logic can be used for  4,4,2 and for 4,3,3

So far there is  3+3+3=9 ways

The only triad not considered so far is 5,2,3

The 5 can be one of 3 colours, the 2 can be one of 2 remaining colours and the 3 must be the last colour. So that is 3*2=6 possibilities.

9+6 = 15 ways that this can be done.

-----------------

Think about it,  see if you can get 36 for the second one by yourself.   

TRY PROOF READING YOUR QUESTIONS AND THEN MAYBE YOU WILL GET SOME ANSWERS!

All you are doing at the moment is wasting your time as well as ours!

Are we supposed to guess the rest of you information?

What is w? 

THANK YOU SO MUCH :)

Earlier request

This question was answered here

No the 4! is not for order mattering. Actually I do not think I have made order matter not within each hand anyway.

The 4! was just for the placement of the cards with 1 on them

The first hand had 4 possibilities, the second hand had 3 possibilities the 3rd hand had 2 possibilities and the last hand took the remaining card.