All Answers 356000 Answers

Number of quarters=N

Number of pennies=1.9N

.25N + .01(1.9N) =8.07, solve for N

N=30 number of quartes

30 x 1.9 =57 number of pennies

30 x .25 + .57 =$8.07

Does a 'penny' means 0.01 dollar and a 'quarter' means 0.25 dollar?

Just a guess......

Let x be the number of quarter he has.

\(0.25x+0.01(1+90\%)x = 8.07\\ 0.25 x + 0.019 x = 8.07\\ 0.269 x = 8.07\\ 269x = 8070\\ x = 30\)

He has 30 quarters.

Looks like I have guessed correctly of the coins' value!! Yay :D

P=314889*e^(0.00644*t) t= 1 to 54 Need to find the value of P

P=2.04... × 10^7

What do you mean? What is that? Post it here and we may be able to help you! :D

When t ranges from 1 to 54,

P ranges from 314889e^(0.00644) to 314889e^(0.34776).

:D

The idea is to get x by itself

(2/3)x<-6    multiply both sides by 3/2

x <  6 (3/2)

x <  18/2

x < 9

OK. :D

"Modulo m graph paper" consists of a grid of m^2 points, representing all pairs of integer residues (x,y) where 0=<x=<m. To graph a congruence on modulo m graph paper, we mark every point (x,y) that satisfies the congruence. For example, a graph of ywould consist of the points (0,0),(1,1) ,(2,4) ,(3,4) , and (4,1).

The graph of  has a single x-intercept $(x_0,0)$ and a single y-intercept $(0,y_0)$, where $0\le x_0,y_0<35$.

What is the value of $x_0+y_0$?

Sorry, too lazy for LaTeX now, 

But....help? I don't understand. :(

Now that I've used a LaTeX box in my reply, the original question seems to be displayed properly.

Weird!!

I don't know how. :(

And this, too.

"Modulo  graph paper" consists of a grid of m^2 points, representing all pairs of integer residues (x,y) where 0=<x=<m. To graph a congruence on modulo m graph paper, we mark every point (x,y) that satisfies the congruence. For example, a graph of y\(\equiv{x^2}\)would consist of the points (0,0),(1,1) ,(2,4) ,(3,4) , and (4,1).

The graph of $$3x\equiv 4y-1 \pmod{35}$$ has a single x-intercept $(x_0,0)$ and a single y-intercept $(0,y_0)$, where $0\le x_0,y_0<35$.

What is the value of $x_0+y_0$?

I suggest you put your LaTeX in the LaTeX box - select the \(\sum \text{LaTeX}\)   Icon on the menu bar of the input page.

You don't need the $$ symbols when you do this.

.

tan(2π/λ-λ/8)= tan π/4 How could this be correct; please explain.

There is more than one answer to this because tan pi/4 = tan 5pi/4 = tan 9pi/4 etc

The most basic answer is found by solving

\(\frac{2\pi}{\lambda}-\frac{\lambda}{8}=\frac{\pi}{4}\\ 8\lambda (\frac{2\pi}{\lambda}-\frac{\lambda}{8})=8\lambda(\frac{\pi}{4})\\ 16\pi-\lambda^2=2\pi \lambda\\ 0=\lambda^2+2\pi \lambda-16\pi\\ \lambda=\frac{-2\pi\pm \sqrt{4\pi^2 +64\pi}}{2}\\ \lambda=\frac{-2\pi\pm 2\sqrt{\pi^2 +16\pi}}{2}\\ \lambda=-\pi\pm \sqrt{\pi^2 +16\pi}\\\)

To get the general solution is the same method just the algebra is a bit worse

\(\frac{2\pi}{\lambda}-\frac{\lambda}{8}=\frac{\pi}{4}+\pi n \qquad n\in Z\\\)

andthen continue the same as the first one.    :)

Junior staff numbers = 28*2/(2+5) → 8

.

HELP!!!

1/ 98 is greater than 1/103 

Answered.

I didn't realize that you didn't actually need an answer Max until I had finished.

I wish you would spell this out in your question so that I (and others) can give my time to people who actually want help.

I am not angry but I do wish you would do as I request. ://

Hi Max :)

\cos^4x-(\sin 2x)(\sin x)(\cos x)-2\cos^2x+\sin^4x+2\sin^2x+1=0

\(\cos^4x-(\sin 2x)(\sin x)(\cos x)-2\cos^2x+\sin^4x+2\sin^2x+1=0\\ \cos^4x+\sin^4x-(\sin 2x)(\frac{sin2x}{2})-2(\cos^2x-\sin^2x)+1=0\\ \cos^4x+\sin^4x+2cox^2xsin^2x-2cox^2xsin^2x-(\sin 2x)(\frac{sin2x}{2})-2(cos2x)+1=0\\ (\cos^2x+\sin^2x)^2-\frac{4cox^2xsin^2x}{2}-(\sin 2x)(\frac{sin2x}{2})-2(cos2x)+1=0\\ 1-\frac{(2cosxsinx)^2}{2}-(\sin 2x)(\frac{sin2x}{2})-2(cos2x)+1=0\\ -\frac{(sin2x)^2}{2}-\frac{(sin2x)^2}{2}-2(cos2x)+2=0\\ -(sin2x)^2-2(cos2x)+2=0\\ -[ 1-cos^2(2x) ]-2(cos2x)+2=0\\ -1+(cos2x)^2 -2(cos2x)+2=0\\ (cos2x)^2 -2(cos2x)+1=0\\ let \;\;y=cos2x\\ y^2-2y+1=0\\ (y-1)^2=0\\ y=1\\ cos2x=1\\ 2x=2\pi n \qquad n\in Z\\ x=\pi n \qquad n\in Z\\ \)

check

\(\cos^4(2\pi n)-(\sin 2*(2\pi n))(\sin (2\pi n))(\cos(2\pi n))-2\cos^2(2\pi n)+\sin^4(2\pi n)+2\sin^2(2\pi n)+1=0\\ \quad 1 \qquad - \qquad 0 \qquad - \qquad 2 \qquad + \qquad 0 \qquad + \qquad 0 \qquad + \qquad 1 \qquad = \qquad 0 \qquad GOOD \)

A.      B.

1.       5.     ncr(6,5) = 6!/(5!1!) → 6

2.       4.     ncr(6,4) = 6!/(4!2!) → 15

3.       3.     ncr(6,3) = 6!/(3!3!) → 20

4.       2.     ncr(6,2) = 6!/(2!4!) → 15

5.       1.     ncr(6,1) = 6!/(1!5!) → 6

Total = 62

.

Hey,

the correct anwser is 62, but thanks lending a helping hand!

A student get 1 coin, B student get 5 coins

A student get 2 coins, B student get 4 coins

A student get 3 coins, B student get 3 coins

A student get 4 coins, B student get 2 coins

A student get 5 coins, B student get 1 coin.

Total 5 ways!! :D

No combination required!!
And if you really need combination for your homework:

6P2 - 6C2 = 5

And I am not sure if this thing above is correct. Sorry for the unsureness(I have made this word up :) hope you understand).

Also, tips!! 

Tips: Please do NOT solve it as a normal trig equation!! It can actually be solved like a quadratic!!

First we find the equation for tangent line;

\(f'(x)=-4x+7\\ \text{Gradient of tangent line}=f'(6)=7-24 = -17\)

When x = 6, f(x)=42 - 2(36)= -30

\(y-(-30)=-17(x-6)\\ y=-17x+72\)

There we go!!

Then we find the gradient for normal.

Negative reciprocal of -17 is 1/17:
Gradient of normal is 1/17

Equation for the normal:

 \(y-(-30)=\dfrac{x-6}{17}\\ y=\dfrac{x-516}{17}\)

Did I make a mistake? This is the first time I know what is normal line. :D

\(\dfrac{2}{3}x-8=0\\ \dfrac{2}{3}x=8\\ x = 8\times \dfrac{3}{2} = 12\)

OR 

If you mean 2/3x - 8 = 0 instead of 2/3 x - 8 = 0...... ('A dot can make extreme difference.' a Chinese idiom translated by me.)

\(\dfrac{2}{3x}-8=0\\ \dfrac{2}{3x}=8\\ \dfrac{1}{x}=8\times \dfrac{3}{2} \\ \dfrac{1}{x}=12\\ x=\dfrac{1}{12}\text{ OR }0.08\bar{3}\text{ <-- the bar is for repeating notation.}\)