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2(10+20) = 60 :))

=^._.^=

Woah, that's an expensive store. 

=^._.^=

The ladder makes a right triangle (since we assume the wall makes a right angle with the ground). Thus the wall and ground are our legs and the ladder is our hypotenuse; try to identify that any time you get these word problems where they're building shapes with objects.

So if we have a triangle with hypotenuse 20 (length of ladder, and a leg of 12 m, what's the angle? We need a trug function... Is that leg opposite or adjacent to the angle? Well the angle is between the ladder and the ground, so it must be adjacent.

What trig function uses the ratio of hypotenuse and adjacent legs? Cosine. So if cos(x) = opp/hy = 12/20 = .6, then all we need is to solve:

X = cos^-1(.6)

(The inverse cosine of .6)

We want to know s - t = ?

We know:

2s = 5t

2s + 5t = 2660

We can substitute in our ratio (first equation) into the second equation twice:

2s + 2s = 2660

5t + 5t = 2660

Solve and find that shirts cost 665, ties cost 266. Put that in our very first question and we get the answer - shop at a different store cause this one's for millionaires (or $399 😅)

Suppose z and w are complex numbers.  
Prove that \(z + \overline{z}\) and \(w \overline{w}\) are real.

\(\begin{array}{|rcll|} \hline \text{Let $z=a+bi$} \\ \text{Let $\overline{z}=a-bi$} \\ \begin{array}{|rcll|} \hline z + \overline{z} &=& a+bi +a-bi \\ \mathbf{z + \overline{z}} &=& \mathbf{2a} \\ \hline \end{array}\\ \\ \text{Let $w=a+bi$} \\ \text{Let $\overline{w}=a-bi$} \\ \begin{array}{|rcll|} \hline w \overline{w} &=& (a+bi)(a-bi) \\ w \overline{w} &=& a^2-(bi)^2 \\ w \overline{w} &=& a^2-b^2i^2 \quad | \quad \mathbf{i^2 = -1} \\ \mathbf{w \overline{w}} &=& \mathbf{a^2+b^2} \\ \hline \end{array}\\ \hline \end{array} \)

After re-reading this question, I realize this is not the same question answered in the link. However, the method for determining the solution is similar. The sample space for the die is notably smaller, (one (1) six in two (2) attempts instead of two (2) sixes in two (2) attempts), while the sample space for the coins remains the same.   

Amy, perhaps you would like to attempt a solution for this after reviewing solution for the more complex dynamic.

Thanks Ginger, and thanks for the compliment.

That was a great interactive thread between the 2 of us :)

Answered here:

Thanks for the link Ginger

Look at the following pattern

I'll leave you to express the resut in the form requested in the question.

[WXYZ] = 18(10*sin45)

thanks asinus :)

Here is the graphs

Let

\(x = 2001^{1002} -2001^{-1002}\) and

\(y = 2001^{1002} + 2001^{-1002}\).

Find \(x^2 - y^2\).

\(\text{Let $a =2001^{1002}$} \\ \text{Let $b =2001^{-1002}$} \\ \text{Let $ab \\\qquad~=2001^{1002} *2001^{-1002}\\\qquad~=2001^{1002-1002}\\\qquad~=2001^0\\\qquad~=1 $} \\ \text{Let $x =a-b$} \\ \text{Let $y =a+b$}\)

\(\begin{array}{|rcll|} \hline x^2-y^2 &=& (a-b)^2-(a+b)^2 \\ x^2-y^2 &=& a^2-2ab+b^2-(a^2+2ab+b^2) \\ x^2-y^2 &=& a^2-2ab+b^2-a^2-2ab-b^2 \\ x^2-y^2 &=& -2ab-2ab \\ x^2-y^2 &=& -4ab \quad | \quad \mathbf{ab=1} \\ x^2-y^2 &=& -4*1 \\ \mathbf{x^2-y^2} &=& \mathbf{-4} \\ \hline \end{array}\)

Can vector AD - CD be simplified?

\(\begin{array}{|rcll|} \hline AD - CD &=& AD + DC \\ \mathbf{AD-CD} &=& \mathbf{AC} \\ \hline \end{array}\)

Hello ellapow!

\(y=\dfrac{x-3}{x-4}\\ x=\dfrac{y-3}{y-4}\\ x(y-4)=y-3\\ xy-4x=y-3\\ xy-y=4x-3\\ y= \dfrac{4x-3}{x-1}\)

\(f^{-1}(x)\ is\ undefined\ at\ x \in\{1\}. \)

Hi Amy,

On the surface, that probability problem seems simple.  The math is simple, but the logic to arrive at the math requires meticulous care.

By Pythagoras' Theorem, we can find side AC

\(AC^2=AB^2+BC^2\)

⇒\(AC^2=25+64\)

\(AC^2=89\)

\(AC=\sqrt {89}\)

So now we know the side of square is \(\sqrt {89}\)

∴ Area  \(=(\sqrt {89})^2\)

            \(=89 sq. units\)

[ABCD] = 60 

AB = x         CD = x + 7           h = 6

1/2(x + x + 7)*6 = 60           x = 6.5

AB = 6.5          CD = 13.5

Median = 1/2(6.5 + 13.5)

1. 16 times, here you go :- 

Number of sides = 360/18

YXW is an interior angle of the 18-gon and its measure is 160º.

Circle radius   R = 25         ( "R" is also the slant height of a cone ) 

Sector arc = 320º

Cone radius   r = 25(320/360)

Cone height    h = sqrt(R² - r²)

Cone volume    V = 1/3(πr²h)

Thanks! This makes more sense now. Now I just need to practice it more before the ap test on tuesday 

We can write -4 in exponential notation as 4e^(pi*i), so the equation is z^4 = 4e^(pi*i).

By Hamilton's Theorem, the solutions are z = 4^{1/4}*e^(pi*i/4), 4^{1/4}*e^(pi*i/4 + pi/4), 4^{1/4}*e^(pi*i/4 + 2*pi/4), and 4^{1/4}*e^(pi*i/4 + 3*pi/4).  Since 4^{1/4} = sqrt(2) and e^(pi*i/4) = (1 + i)/sqrt(2), the first solution is 1 + i.  Then the other roots work out as

4^{1/4}*e^(pi*i/4 + pi/4) = 1 - i,

4^{1/4}*e^(pi*i/4 + 2*pi/4) = -1 - i, and

4^{1/4}*e^(pi*i/4 + 3*pi/4) = -1 + i.