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Your question is not complete. 

You have to tell us what f(x) iexpression is.

Given that k is a positive integer less than 6, how many values can k take on such that 3k=k (mod6)has no solutions in x?

\(\frac{3k}{6}=a+\frac{n}{6}\quad and \quad \frac{k}{6}=b+\frac{n}{6}\quad \\\text{where a, b and n are pos integers and }k<6\)

if k=1   it doesn't work

if k=2   it doesn't work

if k=3    it works

if k= 4  it doesn't work

if k= 5  it doesn't work

so k=3 

k can only take on 1 value and that value is 3

That is ok Rom, I am glad we ended up on the same page :)

How do you get from ( |z| )^2 = a^2 + b^2 to -2z = -2a + 2i *- b

You missed the times sign.

\(z=a+bi\\ \text{|z| is the distance of point Z from the origin.}\\ \text{Using pythagorases theorem we get }|z|^2=a^2+b^2\\ \\~\\ Since z=a+bi\\ -2z=-2(a+bi)\\ -2z=-2a-2bi\)

As follows:

I just lifted everything by one unit and it becomes

Find the volume formed when the region between the graphs  y=7sinx+1   and   y=7cosx+1  for  0<=x<=pi/4

is rotated around the line y=0

\(Volume = \displaystyle \pi\int_0^{\pi/4}\;\left[(7cosx+1)^2-(7sinx+1)^2 \right]dx\\ Volume = \displaystyle \pi\int_0^{\pi/4}\;\left[(49cos^2x+1+14cosx)-(49sin^2x+1+14sinx) \right]dx\\ Volume = \displaystyle \pi\int_0^{\pi/4}\;\left[(49cos^2x-49sin^2x+14cosx-14sinx) \right]dx\\ Volume = \displaystyle \pi\int_0^{\pi/4}\;\left[49(cos^2x-sin^2x)+14cosx-14sinx) \right]dx\\ Volume = \displaystyle \pi\int_0^{\pi/4}\;\left[49(cos2x)+14cosx-14sinx) \right]dx\\ Volume =\pi\left[ \frac{49sin2x}{2}+14sinx+14cosx \right]^{\pi/4}_0\\ Volume =\pi\left[ \frac{49}{2}+\frac{14}{\sqrt2}+\frac{14}{\sqrt2} \right]-\pi\left[ 14 \right]\\ Volume =\pi\left[ \frac{49}{2}+\frac{14\sqrt2}{2}+\frac{14\sqrt2}{2}-\frac{28}{2} \right]\\ Volume =\pi\left[ \frac{41}{2}+\frac{28\sqrt2}{2}\right]\\ Volume =\frac{\pi}{2}\left[ 41+28\sqrt2\right]\quad units^3\)

Here is the original graph

I have lifted everyting by 1 unit to make it easier to work with

1/n + 2/n² = 0.28

Let x = 1/n

Then 2x² + x - 0.28 = 0  which factors as (2x + 1.4)(x - 0.2)

The negative solution is obtained from 2x + 1.4 = 0 or x = - 0.7

hence n = 1/x = -1/0.7  or  n = -10/7 or n ≈ -1.429

ABP = CBD  and EDQ = CDB so, because CBD = CDB we must have ABP = EDQ 

Hence 5x - 7 = 2x + 11  

           3x = 18

             x = 6°

Therefore ABP = 5*6 - 7 or ABP = 23°

ABP + ABD = 180°  so  ABD = 180 - 23  or  ABD = 157°

See heureka's answer #4 here:

https://web2.0calc.com/questions/help-fast-urgent-tommorow-test

Like so:

2. What is the remainder when 5^30 is divided by 7?

5^30 mod 7 =1

If I remember right this question has already been answered. There are no super simple methods.

Anyway you have not said if the boxes are identical or different

 OR if the balls are identical or different.

It makes a BIG difference to the answer.

Four boxes with ball capacity of 3 , 5 , 7 and 8 are given . In how many ways can 19 same balls be put together in these boxes .

Ill assume that the ball are identical and the boxes are identical then you only need to worry about how many balls are in each box. This is not a combination or permutation question it is just a straight counting question.

I'd do it systematically if I were you.

I am almost sure that Heureka has done it in full before.

8 7 1 3

8 7 2 2 

8 6 1 4

8 6 2 3

etc

L:ets try again.

1000 does not have a 4 or a 5 so it is no good.

I want to know how many 3 digit base 7 numbers have at least one 4 or at least one 5

4@@         6       *3     =18

4@#        6*5   *3        =90

44@        6      *3       = 18

444                                 1

total  =  127

same for 5s   = 127

then

454            3

455             3

45*             5*3!=30

total = 36

one or the other or both = 2*127-36 = 218     So 218 have either a 4 or a 5 or both.

There are 343 numbers altogether so that just leaves 125 who do not have either one. 

Hi Rom, I think you have interpreted the question incorrectly.

The question says we must use a 4 or a 5 or both.

If it is only one digit there are only 2 possibilities.

It must be 4 or 5.  So I think that you have answered a different question.

I tried doing it a different way but I still got my original answer. :/

I don't think you got this one quite right Melody.

Consider the 1 digit numbers.  The digit can't be 0, 4, 5.  That leaves 4 possible

2 digit numbers have 4C1 x (5C1) = 20 possibilities

3 digit numbers have 4C1 x (5C1)² = 100 possibilities

There is 1 4 digit number, 1000, and it is counted

This gives 4 + 20 + 100 + 1 = 125 numbers

Oh, duh.  Ok, These are the numbers that don't match what you want.

We have to subtract this from the 343 and we get

343 - 125 = 218

Just like Melody got.  Sorry for the confusion.

How many of the 343 smallest positive integers written in base 7 use 4 or 5 (or both) as a digit?

Ill add the word positive.  Smallest positive integers.

343 base10

7^3=343

so    343 base10     =     1000 base 7

There are a total of 343 numbers.  

Lets think about the base 7 number. And first I am only going take a look at the ones less than 1000 base 7

The 7^2 digit can be 0 to 6 that is 7 digits

The 7^1 digit can be 0 to 6 

The units digit can be 0 to 6 so long as the number is less than 1000 base7

How many have numbers less than 1000 base 7 have NO 4 or 5.

Each digit can be 0,1,2,3 or 6

5*5*5 = 125

[I have actually included 000 and not included 1000 but one in one out makes no difference]

So that is 125  that do not have a 4 or a 5

343- 125 = 218

So I think 218 do have a either a 4 or a 5 or both.

If the digits are all distinct and there are b digits for a 3 digit number we have (we don't start numbers with the digit 0)

(b-1)(b-1)(b-2) = 100

This has 1 real solution, b=6

For number 2 recall Fermat's Little Theorem says \(a^{p-1}\equiv 1 \pmod{p} \text{ for p prime, a a positive integer.}\)

so \(5^6 \equiv 1 \pmod{7}\)

\(5^{30}=\left(5^6\right)^5 \equiv 1^5 \pmod{7} = 1^5 = 1\)

I don't know what the committee forming argument is.

However proving this should be easy enough using the definition

\(k \dbinom{n}{k} = k \dfrac{n!}{k!(n-k)!} = \\ \dfrac{n!}{(k-1)!(n-k)!} = \\ n \dfrac{(n-1)!}{(k-1)!((n-1)-(k-1))!} = \\ n\dbinom{n-1}{k-1}\)

-4(-5-10x) + 12(-2x)
Distribute: 20 + 40x - 24x
Simplify: 16x + 20

5 * (-2) * (-8)
-10 * -8
80 :)

Distance = Speed x Time

Solve for x:
(x + 6) (x^2 - 11 x - 22) = x^2 - 3 x - 54

Expand out terms of the left hand side:
x^3 - 5 x^2 - 88 x - 132 = x^2 - 3 x - 54

Subtract x^2 - 3 x - 54 from both sides:
x^3 - 6 x^2 - 85 x - 78 = 0

The left hand side factors into a product with three terms:
(x - 13) (x + 1) (x + 6) = 0

Split into three equations:
x - 13 = 0 or x + 1 = 0 or x + 6 = 0

Add 13 to both sides:
x = 13 or x + 1 = 0 or x + 6 = 0

Subtract 1 from both sides:
x = 13 or x = -1 or x + 6 = 0

Subtract 6 from both sides:
 x = 13  sub this in Jack's speed:(13^2 - (11*13) - 22) = 4 miles per hour.

Sub 4 into Jill's distance: (13^2  - (3*13) - 54) =76 miles. Divide this by her time:

76 / (13+6) =76/19 =4 miles per hour - Jill's speed = Jack's speed.

Solve for n:
1/n + 2/n^2 = 0.28

0.28 = 7/25:
1/n + 2/n^2 = 7/25

Bring 1/n + 2/n^2 together using the common denominator n^2:
(n + 2)/n^2 = 7/25

Cross multiply:
25 (n + 2) = 7 n^2

Expand out terms of the left hand side:
25 n + 50 = 7 n^2

Subtract 7 n^2 from both sides:
-7 n^2 + 25 n + 50 = 0

The left hand side factors into a product with three terms:
-(n - 5) (7 n + 10) = 0

Multiply both sides by -1:
(n - 5) (7 n + 10) = 0

Split into two equations:
n - 5 = 0 or 7 n + 10 = 0

Add 5 to both sides:
n = 5 or 7 n + 10 = 0

Subtract 10 from both sides:
n = 5 or 7 n = -10

Divide both sides by 7:
 
n = 5     or       n = -10/7

\(x(-4)-6\)

\(x = 7\)

\(7(-4) - 6\)

\(-28 - 6\)

\(-34\)

Evaluates into -34 :)