All Answers 352748 Answers

Thanks Heureka  

Thanks for answering htzhang    

Hi Nirvana,

There was a link so the post was automatically locked.

I have unlocked it.   

Maybe it will help you. 

I am not very good with matrices and cannot answer an open ended question.

Sorry about that.

How many Palindromes between 1,000 and 100,000 are also Prime Numbers?

http://oeis.org A002385 Palindromic primes: prime numbers whose decimal expansion is a palindrome.

...

20 929

21 10301
22 10501
23 10601
24 11311
25 11411
26 12421
27 12721
28 12821
29 13331
30 13831
31 13931
32 14341
33 14741
34 15451
35 15551
36 16061
37 16361
38 16561
39 16661
40 17471
41 17971
42 18181
43 18481
44 19391
45 19891
46 19991
47 30103
48 30203
49 30403
50 30703
51 30803
52 31013
53 31513
54 32323
55 32423
56 33533
57 34543
58 34843
59 35053
60 35153
61 35353
62 35753
63 36263
64 36563
65 37273
66 37573
67 38083
68 38183
69 38783
70 39293
71 70207
72 70507
73 70607
74 71317
75 71917
76 72227
77 72727
78 73037
79 73237
80 73637
81 74047
82 74747
83 75557
84 76367
85 76667
86 77377
87 77477
88 77977
89 78487
90 78787
91 78887
92 79397
93 79697
94 79997
95 90709
96 91019
97 93139
98 93239
99 93739
100 94049
101 94349
102 94649
103 94849
104 94949
105 95959
106 96269
107 96469
108 96769
109 97379
110 97579
111 97879
112 98389
113 98689

114 1003001

...

93 Palindromes are between 1,000 and 100,000

Which is larger, the blue area or the orange area?

\(\begin{array}{|rcll|} \hline {\color{orange}\text{orange}} +3\times {\color{blue}\text{blue}} &=& \pi r_{\text{circumcircle}}^2 \quad | \quad : {\color{orange}\text{orange}} \\ 1+3\times \dfrac{ {\color{blue}\text{blue}} } { {\color{orange}\text{orange}} } &=& \dfrac{ \pi r_{\text{circumcircle}}^2 } { {\color{orange}\text{orange}} } \quad | \quad {\color{orange}\text{orange}} = \pi r_{\text{incircle}}^2 \\ 1+3\times \dfrac{ {\color{blue}\text{blue}} } { {\color{orange}\text{orange}} } &=& \dfrac{ \pi r_{\text{circumcircle}}^2 } { \pi r_{\text{incircle}}^2 } \\ 1+3\times \dfrac{ {\color{blue}\text{blue}} } { {\color{orange}\text{orange}} } &=& \left( \dfrac{ r_{\text{circumcircle}} } { r_{\text{incircle}} } \right)^2 \\ 3\times \dfrac{ {\color{blue}\text{blue}} } { {\color{orange}\text{orange}} } &=& \left( \dfrac{ r_{\text{circumcircle}} } { r_{\text{incircle}} } \right)^2 -1 \\ \dfrac{ {\color{blue}\text{blue}} } { {\color{orange}\text{orange}} } &=& \dfrac{1}{3}\left( \left( \dfrac{ r_{\text{circumcircle}} } { r_{\text{incircle}} } \right)^2 -1 \right) \\ \\ && \boxed{\text{here, if triangle is equilateral: }\\ \mathbf{2\times r_{\text{incircle}} = r_{\text{circumcircle}}!!!} } \\\\ \dfrac{ {\color{blue}\text{blue}} } { {\color{orange}\text{orange}} } &=& \dfrac{1}{3}\left( \left( \dfrac{ 2\times r_{\text{incircle}} } { r_{\text{incircle}} } \right)^2 -1 \right)\\ \dfrac{ {\color{blue}\text{blue}} } { {\color{orange}\text{orange}} } &=& \dfrac{1}{3}\times(2^2 -1) \\ \dfrac{ {\color{blue}\text{blue}} } { {\color{orange}\text{orange}} } &=& \dfrac{1}{3}\times(3) \\ \dfrac{ {\color{blue}\text{blue}} } { {\color{orange}\text{orange}} } &=& 1 \\ \mathbf{ {\color{blue}\text{blue}} } &=& \mathbf{ {\color{orange}\text{orange}} } \\ \hline \end{array}\)

What are the two smallest prime factors of \(2^{1024} - 1\)?

\(\begin{array}{|rcll|} \hline && 2^{1024} - 1 \\ &=& \left( 2^{512} - 1\right) \left( 2^{512} + 1\right) \\ &=& \left( 2^{256} - 1\right) \left( 2^{256} + 1\right) \left( 2^{512} + 1\right) \\ &=& \left( 2^{128} - 1\right) \left( 2^{128} + 1\right) \left( 2^{256} + 1\right) \left( 2^{512} + 1\right) \\ &=& \left( 2^{64} - 1\right) \left( 2^{64} + 1\right) \left( 2^{128} + 1\right) \left( 2^{256} + 1\right) \left( 2^{512} + 1\right) \\ &=& \left( 2^{32} - 1\right) \left( 2^{32} + 1\right) \left( 2^{64} + 1\right) \left( 2^{128} + 1\right) \left( 2^{256} + 1\right) \left( 2^{512} + 1\right) \\ &=& \left( 2^{16} - 1\right) \left( 2^{16} + 1\right) \left( 2^{32} + 1\right) \left( 2^{64} + 1\right) \left( 2^{128} + 1\right) \left( 2^{256} + 1\right) \left( 2^{512} + 1\right) \\ &=& \left( 2^{8} - 1\right) \left( 2^{8} + 1\right) \left( 2^{16} + 1\right) \left( 2^{32} + 1\right) \left( 2^{64} + 1\right) \left( 2^{128} + 1\right) \left( 2^{256} + 1\right) \left( 2^{512} + 1\right) \\ &=& \left( 2^{4} - 1\right) \left( 2^{4} + 1\right) \left( 2^{8} + 1\right) \left( 2^{16} + 1\right) \left( 2^{32} + 1\right) \left( 2^{64} + 1\right) \left( 2^{128} + 1\right) \left( 2^{256} + 1\right) \left( 2^{512} + 1\right) \\ &=& \left( 2^{2} - 1\right) \left( 2^{2} + 1\right) \left( 2^{4} + 1\right) \left( 2^{8} + 1\right) \left( 2^{16} + 1\right) \left( 2^{32} + 1\right) \left( 2^{64} + 1\right) \left( 2^{128} + 1\right) \left( 2^{256} + 1\right) \left( 2^{512} + 1\right) \\ &=& \left( 2^{1} - 1\right) \left( 2^{1} + 1\right) \left( 2^{2} + 1\right) \left( 2^{4} + 1\right) \left( 2^{8} + 1\right) \left( 2^{16} + 1\right) \left( 2^{32} + 1\right) \left( 2^{64} + 1\right) \left( 2^{128} + 1\right) \left( 2^{256} + 1\right) \left( 2^{512} + 1\right) \\ &=& 1\times 3 \times 5 \times \left( 2^{4} + 1\right) \left( 2^{8} + 1\right) \left( 2^{16} + 1\right) \left( 2^{32} + 1\right) \left( 2^{64} + 1\right) \left( 2^{128} + 1\right) \left( 2^{256} + 1\right) \left( 2^{512} + 1\right) \\ \hline \end{array} \)

The two smallest prime factors are 3 and 5, they are also the smallest possible odd prime numbers,

all factors are odd,

so that the 2 as the only even prime number cannot be included

Here is a more geometric approach. I guess you can call mine Euclidean, whereas our moderator's approach is Cartesian.

Let the line segment HF be perpendicular to CB. Then from the fact that BHF and BEC are similar triangles we can deduce

\(\frac{HF}{2}=\frac{4}{7}\)and therefore \(HF=\frac{8}{7}\). From the similarity of triangles AED and FHD and the fact that the ratio of the sides is

\(\frac{8}{7}\div2=\frac{4}{7}\) we can conclude that \(c=(\frac{4}{7})^2\cdot b=\frac{16}{49}\cdot b\) (lower case letters indicate areas of the respective regions). Furthermore, from the fact that \((c+e) + a=7\ and \ b+a=6\) we can conclude that \((c+e)-b=1\). We also know that \(e=2\cdot\frac{8}{7}=\frac{16}{7}\).Therefore,

\((\frac{16}{49}\cdot b+\frac{16}{7})-b=1\) and this implies that \(b=\frac{21}{11}\). Since \(d+b=14-7=7 \), we have \(d=7-\frac{21}{11}=\frac{56}{11}\), the looked-for area opf the triangle ADB.

In the figure, what is the area of triangle \(ABD\)? Express your answer as a common fraction.

awesomey sauce! thanks again :D

yes yes!! thank you so much CPhill!

-2f(x)   vertically stretches the graph by a factor of 2 and  "flips" it over the x axis

f( -2x)  horizontally compresses the graph    by a factor of 2 (makes it narrower) and "flips" it across the y axis

f(-1/2 x)  horizontally compresses the graph by a factor of 1/2 (makes it wider) and "flips" it acroos the y axis  

See the last two results here for  f(x)  = (x - 3)^2  : https://www.desmos.com/calculator/ql7h8ppdc5

See if this helps :

So    a 90°  counter-clockwise rotation  would put the vector  pointing up the y axis in a positive direction

The rotational matrix for  a 90° counter-clockwise rotation is the one you were given 

But....this is the same thing as a 270°  clockwise rotation

  Does this make sense  ???

The point \((r,\theta)\) in polar coordinates is \((7,5)\) in rectangular coordinates.
What is the point \(\left( 2r, \theta + \dfrac{\pi}{2} \right)\)  in rectangular coordinates?

\(\begin{array}{|rcll|} \hline && \text{Set $(7,5)$ to $7+5i$ } \\\\ && \text{Rotation by } \dfrac{\pi}{2}: \qquad \times i \\ && \text{scaling} : \qquad \times 2 \\\\ && 2\times(7+5i)\times i \\ &=& 2\times(7i+5i^2) \quad | \quad i^2 = -1 \\ &=& 2\times(7i-5) \\ &=& -10+14i \\\\ && \mathbf{\left( 2r, \theta + \dfrac{\pi}{2} \right) = (-10,14)} \\ \hline \end{array}\)

on my paper i said 90 degrees counterclockwise which.............i thought that 90 degrees clockwise meant 270 degrees ccw and -90 meant 270 ccw 

woah....what im so confused im so sorry )::::

Your answer was incorrect

A 90 °  rotation clockwise  about the origin  has  the tranformation formula :

(x, y) ⇒ (  y, -x)

So

(5, -2)  would become ( -2, -5)

(3,2)  would become (2, -3)

(0,0)  would become (0, 0)

(-2, -4) would become ( -4, 2)

But your  transformed  points should be for a -90°  rotation  [ =  a 270°  rotation ]

(2,5)   (-2,3)  (0, 0)   and  ( -4, -2)

Which  are the points on your transformed matrix

ohhhhhh i see! 

kind of like stretches and compressions! 

1/2f(X) vertically compresses the graph
but 

f(2x) horizontally compresses the graph as well

and 2f(X) vertically stretches 
as f(1/2x) horizontally stretches 

but....
what would happen with -2f(x) 
and f(-2x) and f(-1/2x)

This is always confusing

f ( x- a)    shifts the graph of  f(x)   "a"  units to the right

f( x + a)   shifts the graph of f(x)  "a" units to the left

Example

f(x)  = x^2      has a vertex at   (0, 0)

But

f(x) = (x -3)^2   has a vertex at  (3, 0)      .......3 units to the right of the original vertex

f(2.6)  =  4(2.6)^2  - 10(2.6) + 9  = 10.04

f(5.2)  = 4(5.2)^2 - 10(5.2) + 9  =65.16

So    we have the points   ( 2.6, 10.04)  and  ( 5.2, 65.16)

The slope is

[ 65.16 - 10.04 ]  / [ 5.2 - 2.6 ]   = 21.2  = avg rate of change

oh! 
wait so now im confused........am i right or wrong.....

The matrix for  a 90°  counter-clockwise rotation  is

[ 0   -1  ]

[ 1    0  ]

But this is the same as a 270° clockwise rotation

See the rotation matrices here : https://en.wikipedia.org/wiki/Rotation_matrix

im not the person who asked this question but, why do we add 3 and not subtract if it says -3............. i know based off of f(x)=(x-h) that the function should move h units to the right,,,,but im confused on why we add and not subtract

OK. Thank you, but I have a question.......since they told us to multiply by \(\begin{bmatrix} 0&& -1 \\ 1&& 0 \end{bmatrix}\)

shouldn't it be 90 degrees? 

Uhhh I think I did something wrong :PP I got (15-5)\(3-2) = 10 as the slope but I don't think that's right.. :((

2^1024 -1 = 3 x 5 x 17 x 257 x 641 x 65537 x 274177 x 2424833 x 6700417 x 67280421310721.........etc

Note that  the points (-2,3) and (1, 3)  are on h(x)

h (x - 3)  shifts  the graph of h(x)  3 units to the right...so  that the point (-2, 3)  becomes (1,3)

So...the two graphs will intersect at (1, 3)

The sum of the coordinates =  1 + 3   =   4