JKP1234567890

http://mrswisey.weebly.com/uploads/2/3/3/9/23393696/d3-_arithmetic_sequences_answer_key_all.pdf 

scroll down to question 10 (I think it's the same question)

I don't know if you meant 3 times 3 or $3x^3$? 

please tell me so i can help you solve the problem

combos: 

(0,0,25)

(0,1,20)

(0,2,15)

(0,3,10)

(0,4,5)

(0,5,0)

(1,0,15)

(1,1,10)

(1,2,5)

(1,3,0)

(2,0,5)

(2,1,0)

12 ways

d=number of domestic stamps

f=number of foreign stamps 

$\:\frac{d}{f}=\frac{5}{2}$

$2d=5f$

$\frac{d}{f+12}=\frac{2}{1}$

$d=2f+24$

substitute d in first equation: 

$2(2f+24)=5f$

$f=48$

$d=120$

JP

(x+30)+(x+40)+(2x+10)=180

4x+80=180

4x=100

x=25

JP

Well, lets see, for an equation to have more than one solution, even if that box has a "y" in it, then it still produces more than one solution (it's a quadratic equation). I don't know if you mean if it has infinite solutions, in that case, it would be: 

$\frac{y}{2}+\frac{1}{4}=3+xy$

$2y+1=12+4xy$

$2y-11=4xy$

$x=\frac{2y-11}{4y\:}$  

unless, of course y=0, since anything divided by 0 is undefined         

JP 

I'm sorry, can you tell me how I can delete my solution? (since after people have replied to your post, I can't find a way to delete it)

Here's the explanation:

I couldn't find a picture, but the right angle corner is point C.  

So, C's coordinates are $\left(x_2,\:y_1\right)$

BC is $\left(y_2-y_1\right)$

AC is $\left(x_2-x_1\right)$

so, $AB^2=AC^2+BC^2=\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2$

$\:AB=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$

because of DeMoivre's Theorem:

$\begin{align*} \operatorname{cis} n \theta &= (\operatorname{cis} \theta)^n \\ &= (\cos \theta + i \sin \theta)^n \\ &= \cos^n \theta + \binom{n}{1} i \cos^{n - 1} \theta \sin \theta - \binom{n}{2} \cos^{n - 2} \theta \sin^2 \theta - \binom{n}{3} i \cos^{n - 3} \theta \sin^3 \theta + \dotsb. \end{align*}$

$\begin{align*} \cos n \theta &= \cos^n \theta - \binom{n}{2} \cos^{n - 2} \theta \sin^2 \theta + \binom{n}{4} \cos^{n - 4} \theta \sin^4 \theta - \dotsb, \\ \sin n \theta &= \binom{n}{1} \cos^{n - 1} \theta \sin \theta - \binom{n}{3} \cos^{n - 3} \theta \sin^3 \theta + \binom{n}{5} \cos^{n - 5} \theta \sin^5 \theta - \dotsb. \end{align*}$

$\begin{align*} \sin 5 \theta &= \binom{5}{1} \cos^4 \theta \sin \theta - \binom{5}{3} \cos^2 \theta \sin^3 \theta + \binom{5}{5} \sin^5 \theta \\ &= 5 \cos^4 \theta \sin \theta - 10 \cos^2 \theta \sin^3 \theta + \sin^5 \theta. \end{align*}$

$\sin 5 \theta = \sin^5 \theta$ 

becomes 

$5 \cos^4 \theta \sin \theta - 10 \cos^2 \theta \sin^3 \theta + \sin^5 \theta = \sin^5 \theta$

$5 \cos^4 \theta \sin \theta - 10 \cos^2 \theta \sin^3 \theta = 0$

$\theta$ is an acute angle, so $\cos^2 \theta - 2 \sin^2 \theta = 0$

$\cos^2 \theta = 2 \sin^2 \theta$

so, I know that $\tan^2 \theta = \frac{1}{2}$

$\tan \theta = \frac{1}{\sqrt{2}}$

$\tan 2 \theta = \frac{2 \tan \theta}{1 - \tan^2 \theta} = \frac{\sqrt{2}}{1 - \frac{1}{2}} = {2 \sqrt{2}}$

226 numbers between 25 and 250 inclusive. 

11 squares between 25 and 250.

4 cubes between 25 and 250.

only perfect square and perfect cube between 25 and 250 is 64. 

11+4-1=14

226-14=212