All Answers 352833 Answers

this is \(H_{1000}\), the 1000th harmonic number.

It is approximately 7.48547

It is exactly

(5336291328229478504559104562404298040965247228038426009710134924845626\
8889497101757506097901985035691409088731550468098378442172117885009464\
3023443265660225021002784256328520814055449412104425101426727702947747\
1270891796396777961045322469242686646888828158207198489710511079687324\
9319155529397017508931564519976085734473014183284011724412280649074307\
7037366831700558002936592350885893602352858528081607595747378366554131\
75508131522517/)\
(7128865274665093053166384155714272920668358861885893040452001991154324\
0875811114994764441519138715869117178170195752565129802640676210092514\
6587100430513107268626814320019660997486274593718834370501543445252373\
9745298963145674982128236956232823794011068809262317708861979540791247\
7545580493264757378299233527517967352480424636380511370343312147817468\
5087845348567802188807537324992199567205693202909939089168748767269795\
0931603520000)

thank you

my attempt:

apple = a, pear = p

5a = 6p

a = p+0.15

5(p+0.15) = 6p

5p + 0.75 = 6p

0.75 = p

0.75+ 0.15 = 0.9

0.9 = a

5(0.9) + 6( 0.75) = £9

\(\text{the curve passes through the given point }(1,3)\\ 2(1)^2-3(1)+c = 3\\ 2-3+c=3\\ c=4\\ f(x) = 2x^2-3x+4\)

If the answer is not 15 then there must be some restriction on how many stickers must be on each piece of paper.

If papers are allowed to have 0-8 stickers then there are 15 ways to place them if only the number of stickers per page matters.

Maybe the pieces of paper are actually distinct, you just don't care about which stickers are on them.

In this case there will be \(\dbinom{11+3-1}{3-1} = \dbinom{11}{3}=165\) different arrangements.

this is my attempt:

let the 7 consecutive number be n, n+1, n+2, n+3, n+4, n+5, n+6

add them all together u have 7n+21

which means 7n+21 = 182

from the system we can see n is our aim as it is the smallest number 

7n = 182-21

7n= 161

n =161/7

n=23 

so the answer is 23

hope this helps

63^2 = 3,969

There is a more formal method to solving this problem.

First let's reduce these congruences.

\(7x \equiv 21 \pmod{14}\\ x \equiv 3 \pmod{2}\\ x \equiv 1 \pmod{2}\)

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

\(2x+13 \equiv 16 \pmod{9}\\ 2x \equiv 3 \pmod{9}\\ \text{now we multiply both sides by the multiplicative inverse of }2 \pmod{9}\\ 5\cdot 2x \equiv 5\cdot 3 \pmod{9}\\ x \equiv 6 \pmod{9}\)

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

\(-2x + 1 \equiv x \pmod{25}\\ 3x \equiv 1 \pmod{25}\\ 17\cdot 3x \equiv 17 \pmod{25}\\ x \equiv 17 \pmod{25}\)

\(\text{Now we have the following system of linear congruences}\\ x \equiv 1 \pmod{2}\\ x \equiv 6 \pmod{9}\\ x \equiv 17 \pmod{25}\)

\(\text{From the first congruence we have}\\ x = 2t+1,~t \in \mathbb{Z}\\ \text{substitute this into the second congruence}\\ 2t+1 \equiv 6 \pmod{9}\\ 2t\equiv 5 \pmod{9}\\ 5\cdot 2t \equiv 5\cdot 5 \pmod{9}\\ t \equiv 7 \pmod{9}\\ t = 9s+7,~s \in \mathbb{Z}\)

\(\text{substitute this back into }x \text{ and simplify}\\ x = 2(9s+7)+1\\ x = 18s + 14+1 = 18s+15\)

\(\text{Now we substitute this into the third congruence}\\ 18s+15 \equiv 17 \pmod{25}\\ 18s \equiv 2 \pmod{25}\\ \text{Now we have to find }18^{-1} \pmod{25},\text{ with these small numbers trial and error works}\\ 18^{-1}\pmod{25} = 7\\ 7\cdot 18 s \equiv 14 \pmod{25}\\ s \equiv 14 \pmod{25}\\ s = 25u+14,~u\in \mathbb{Z}\)

\(\text{and we substitute this back into }x\\ x = 18(25u+14)+15 = 267+450u\\ x \equiv 267 \pmod{450}\\ \text{The smallest 4 digit solution will be}\\ x = 2\cdot 450 + 267 = 1167\)

asdf335:    I think you have made several mistakes in your reasoning.

1 - 8 + 7 + 6 + 5 +........+ 1 = 36 not 28

2- You should remove all numbers beginning with zeros to the right of 9, because they repeat: E.G. 9 011 2, 9 022 2, 9 033 2.......etc. There are 9 in total.

3- For each subsequent number beginning with 1, there are about 2 numbers repeated, which should also be removed: E.G. 9 112 2, 9 122, 2. There are 2 such numbers in each category beginning with 1 and ending in 9.  The only exceptions are the numbers beginning with 5: 95062, 95172, 95282, 95392, 95502, 95612, 95722, 95832, 95942. Out of these 9 numbers, there is one repeat, i.e,: 9 550 2, which should be removed.

4- There are a total of 9 x 10 =90 - 9 zeros - (2 x 9) + 1 for the other 9 numbers = 64 combinations of the 3 numbers a, b, c.

5- Here are All the numbers that meet the condition that they be "distinct", or different from each other:

91322, 91432, 91542, 91652, 91762, 91872, 91982, 92092, 92312, 92532, 92642, 92752, 92862, 92972, 93082, 93192, 93412, 93522, 93742, 93852, 93962, 94072, 94182, 94292, 94512, 94622, 94732, 94952, 95062, 95172, 95282, 95392, 95612, 95722, 95832, 95942, 96052, 96272, 96382, 96492, 96712, 96822, 96932, 97042, 97152, 97262, 97482, 97592, 97812, 97922, 98032, 98142, 98252, 98362, 98472, 98692, 98912, 99022, 99132, 99242, 99352, 99462, 99572,  99682=64

\(\text{You start your answer with an assertion that needs proving}\\ 9abc2 \equiv 0 \pmod{11} \Rightarrow (9-a+b+c+2) \equiv 0 \pmod{11}\)

\(90000 + 1000a + 100b+10c+2 \equiv 0 \pmod{11}\\ 9 + 10a + b +10c+2 \pmod{11} = \\ 11-a+b-c \pmod{11} = \\ b - (a+c) \pmod{11}\)

Neat little trick.

Handy: 75+45(6)=345

Fixit: 135+50(6)=300+135=435.

435-345=90.

yes i have, thank you for your help

\(\bar{x} = \dfrac{1}{3-0} \displaystyle \int \limits_0^3~2x^2-3~dx = \\ \dfrac 1 3 \left(\left. \dfrac 2 3 x^3 - 3x \right |_0^3\right) = \\ \dfrac 1 3 \left(18-9\right) = \\ \dfrac 9 3 = 3\)

I suspect you omitted the factor of 1/3 used to normalize the integral

I assume you can do part (b)

using your other diagram, take the midpoint of AB as F, and draw line segment OF. now you have 15-75-90 triangle, and it has hypotenuse 1.

because sin 15 is (sqrt6-sqrt2)/4 and cos 15 is (sqrt6+sqrt2)/4, you just multiply these values to get 1/4. now you divide by 2 because of triangle area, so it is 1/8. there are 12*2=24 of these triangles, so the answer is 24*1/8 or 3.

for the second one, this means the legs of the right triangles are 2sqrt6, and the hypotenuse is 4sqrt3. the hypotenuse is also the same as the side length of the square, so the answer is 48.

for the third one, i am not sure, but i think you need to use the fact that sin(2x)=2 sin x * cos x.

for the fourth one, triangle BFD trisects line AE with some intuition.

because line AE is sqrt3, the side length of the smaller hexagon is sqrt3/3. using the formula 3s^2sqrt3/2 for the area of a hexagon, you get sqrt3/2.

i will leave the last one for you to do because i do not know how to do it.

HOPE THESE ANSWERS ALL HELPED!!!!!!!!!!!!!

Thanks so much! Now I understand it!!

this means that 9-a+b-c+2 is divisible by 11, so b-(a+c) is divisible by 11.

so this value is either -11 or 0.

if it is -11, we can use casework.

first take the case where b is 0.

the a can be 2, c can be 9, or a can be 3, and c can be 8, all the way up to the case where a is 9 and c is 2.

this has 8 possibilities.

now you take the case where b is 1. the sums are 12, and there are 7 posibilities using the same logic.

you keep doing this untill you get 1 possibility, so there are 8+7+6+5+...+1 possibilities or 28.

now if b-(a+c)=0, then it is the same.

if b is 0, there is one possibility.

if b is 1 there are 2 possibilities.

and you keep doing this until b is 9 and there are 10 possibilities.

this makes 1+2+3+...+10=55 possibilities for that case.

now you just add them up and get 55+28=83 possibilities.

HOPE THIS HELPED!!

because there are twelve sides, you get twelve secotrs with 30 degrees each.

this means triangle AMO is 30-60-90 with a hypoutenuse of 1.

this means AM is 1/2.

HOPE THIS HELPED!

this means they are either both positive, both negative, or x^2+x+3 is 0.

x^2+x+3 can not be 0 because of the quadratic formula, so they are either both positive or both negative.

if they are both positive, then 2x^2+x-6 should be positive. factoring, it is (2x-3)(x+2), so the roots are 3/2 and -2.

since this is positive, x is greater than 3/2 or less than -2. in these cases, x^2+x+3 will be positive (plug in some numbers).

now you need to do the case where they are both negative.

in this case, x is between -2 and 3/2 in the denominator.

now you try plugging in values. the +3 in the numerator will overpower any value in there so it will not be negative, so we can rule this case out.

so the answer is \(\boxed{x<-2}\) or \(\boxed{x>{3\over2}}\).

HOPE THIS HELPED!!

Down voting a guest is a symbolic statement, Tertre.

In this case it’s also a catalytic condiment for a symbolic and hardy, “STFU!”

I just want to see how other users tackle this question, sorry for being rude before.

tertre you have to do it again with the same logic and get 722.

lol :P

pleez tertre give us a good explanation or something

Sorry. Nice explanation, asdf!

Sorry, I just want to see cphill's answer. I want to see his solution.

thanks melody for the advice!