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Right triangle    distance apart = hypotenuse = c

Pythagorean Theorem:

c^2 = 12^2 + 5^2      solve for  'c' 

As follows:

Find the area of the square.

\(\text{Let the side of the square $=a$} \\ \text{Let the area of the square $=a^2$} \\ \text{Let the radius of the circle $=r$} \\ \text{Let the center of the circle $(x_c,y_c)=(-r,\ 0)$} \)

circle:

\(\begin{array}{|rcll|} \hline \mathbf{\text{circle:}} \\ \hline \mathbf{(x+r)^2+y^2} &=& \mathbf{r^2} \quad | \quad x=-\dfrac{a}{2} \\\\ \left(r-\dfrac{a}{2}\right)^2+y^2 &=& r^2 \\\\ r^2 -2\cdot \dfrac{ar}{2}+\dfrac{a^2}{4} +y^2 &=& r^2 \\\\ r^2 - ar +\dfrac{a^2}{4} +y^2 &=& r^2 \\ y^2 &=& ar -\dfrac{a^2}{4} & \boxed{2r = a-6 \\ r=\dfrac{a-6}{2}} \\ y^2 &=& a\left(\dfrac{a-6}{2}\right) -\dfrac{a^2}{4} \\\\ y^2 &=& 2a\dfrac{(a-6)}{4} -\dfrac{a^2}{4} & \boxed{4+y = \dfrac{a}{2}\\y=\dfrac{a}{2}-4\\y=\dfrac{a-8}{2} } \\ \left(\dfrac{a-8}{2}\right)^2 &=& 2a\dfrac{(a-6)}{4} -\dfrac{a^2}{4} \\\\ \dfrac{(a-8)^2}{4} &=& 2a\dfrac{(a-6)}{4} -\dfrac{a^2}{4} \\\\ (a-8)^2 &=& 2a(a-6) -a^2 \\ a^2-16a+64 &=& 2a^2-12a-a^2 \\ -16a+64 &=& -12a \\ 64 &=& 4a \quad | \quad :4 \\ 16 &=& a \\ \mathbf{a} &=& \mathbf{16} \\ \hline \end{array}\)

The area of the square is \(a^2 = 16^2 = \mathbf{256}\)

Plugging this into a calculator: (2000/2)4.5 + (3000/2000)40 + (18 * 3000) = 58560.

WOW thx so Much!! I like the way you explaned it step by step! Thank you agian :D

1.

Find the number of ordered pairs (a,b) of integers such that \(\mathbf{\dfrac{a + 2}{a + 5} = \dfrac{b}{3}}\)

\(\begin{array}{|rcll|} \hline \mathbf{\dfrac{a + 2}{a + 5}} &=& \mathbf{\dfrac{b}{3}} \\\\ 3(a+2) &=& b(a+5) \\ 3a+6 &=& ab+5b \\ -3a+5b+ab &=& 6 \\ -3a+5b+ab+(-3*5) &=& (-3*5)+6 \\ -3a+5b+ab-15 &=& -15+6 \\ \mathbf{(a+5)(b-3)} &=& \mathbf{-9} \\ \mathbf{m*n} &=& \mathbf{-9} \\ && \text{find all integer products }\ m*n = -9, \\ && \text{where}\ m=a+5,\ \text{ and }\ n=b-3 \\ \hline \end{array}\)

all integer products

\(\begin{array}{|ccl|r|r|r|r|} \hline && m*n & m & n & a=m-5 & b=n+3 \\ \hline -9 &=& 1\times -9 & 1 & -9 & -4 & -6 \\ -9 &=& -1\times 9 & -1 & 9 & -6 & 12 \\ -9 &=& 3\times -3 & 3 & -3 & -2 & 0 \\ -9 &=& -3\times 3 & -3 & 3 & -8 & 6 \\ -9 &=& 9\times -1 & 9 & -1 & 4 & 2 \\ -9 &=& -9\times 1 & -9 & 1 & -14 & 4 \\ \hline \end{array}\)

There are 6 integer products.
The number of ordered pairs (a,b) are 6.
\((-4,-6),\ (-6,12),\ (-2,0),\ (-8,6),\ (4,2),\ (-14,4)\)

Suppose m = 2 + 6i, and |m+n| = 3sqrt10, where n is a complex number.
a. What is the minimum value of the modulus of n?
b. Provide one example of the complex number, n.

My attempt:

\(\begin{array}{|rcll|} \hline m&=& 2+6i \\ n&=& x+yi \\ \hline m+n &=& (2+x) + (6+y)i \\ \mathbf{|m+n|^2} = (2+x)^2 + (6+y)^2 &=& (3\sqrt{10})^2 \\ (2+x)^2 + (6+y)^2 &=& 9\cdot 10 \\ (2+x)^2 + (6+y)^2 &=& 90 \\ \hline \mathbf{|n|^2} &=& x^2+y^2 \\ \hline \end{array}\)

We wish to minimize \( f ( x , y ) = x^2 + y^2\)
subject to the constraint \((2+x)^2+(6+y)^2=90\)

For the method of Lagrange multipliers, the constraint is \(g(x,y) =(2+x)^2+(6+y)^2 - 90 \)

hence
\(\Lambda(x,y,\lambda) = f(x,y)+\lambda g(x,y) \\ \Lambda(x,y,\lambda)= x^2 + y^2+\lambda\left[(2+x)^2+(6+y)^2 - 90\right] \)

Now we can calculate the gradient:

\(\begin{array}{|rcll|} \hline \dfrac{\partial \Lambda}{\partial x} &=& 2x+2\lambda(2+x) \\ \hline \end{array} \begin{array}{|rcll|} \hline \dfrac{\partial \Lambda}{\partial y} &=& 2y+2\lambda(6+y) \\ \hline \end{array} \begin{array}{|rcll|} \hline \dfrac{\partial \Lambda}{\partial \lambda} &=& (2+x)^2+(6+y)^2 - 90 \\ \hline \end{array} \)

and therefore:

\(\begin{array}{|rcll|} \hline 2x+2\lambda(2+x) &=& 0 \quad | \quad : 2 \\ x+\lambda(2+x) &=& 0 \\ x+2\lambda+\lambda x &=& 0 \\ x(1+\lambda) &=& -2\lambda \\ \mathbf{x} &=& \mathbf{-\dfrac{ 2\lambda}{1+\lambda} } \\ \hline \end{array} \begin{array}{|rcll|} \hline 2y+2\lambda(6+y) &=& 0 \quad | \quad : 2 \\ y+\lambda(6+y) &=& 0 \\ y+6\lambda+\lambda y &=& 0 \\ y(1+\lambda) &=& -6\lambda \\ \mathbf{y} &=& \mathbf{-\dfrac{ 6\lambda}{1+\lambda} } \\ \hline \end{array}\\ \begin{array}{|rcll|} \hline (2+x)^2+(6+y)^2 - 90 &=& 0 \\ (2+x)^2+(6+y)^2 &=& 90 \\ \left( 2-\dfrac{ 2\lambda}{1+\lambda} \right)^2+\left( 6-\dfrac{ 6\lambda}{1+\lambda} \right)^2 &=& 90 \\ \dfrac{ (2+2\lambda -2\lambda)^2 + (6+6\lambda-6\lambda)^2}{(1+\lambda)^2} &=& 90 \\ \dfrac{2^2+6^2}{(1+\lambda)^2} &=& 90 \\ \dfrac{40}{(1+\lambda)^2} &=& 90 \\ \ldots \\ (1+\lambda)^2 &=& \dfrac{40}{90} \\ (1+\lambda)^2 &=& \dfrac{4}{9} \\ 1+\lambda &=& \dfrac{2}{3} \\ \lambda &=& \dfrac{2}{3}-1 \\ \mathbf{\lambda} &=& \mathbf{-\dfrac{1}{3}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{x} &=& \mathbf{\dfrac{-2\lambda}{1+\lambda} } \quad | \quad \mathbf{\lambda=-\dfrac{1}{3}} \\\\ x &=& \dfrac{ -2\left(-\dfrac{1}{3}\right)}{1-\dfrac{1}{3}} \\\\ x &=& \dfrac{ \dfrac{2}{3}}{\dfrac{2}{3}} \\\\ \mathbf{x} &=& \mathbf{1} \\ \hline \end{array} \begin{array}{|rcll|} \hline \mathbf{y} &=& \mathbf{\dfrac{-6\lambda}{1+\lambda} } \quad | \quad \mathbf{\lambda=-\dfrac{1}{3}} \\\\ y &=& \dfrac{ -6\left(-\dfrac{1}{3}\right)}{1-\dfrac{1}{3}} \\\\ y &=& \dfrac{ \dfrac{6}{3}}{\dfrac{2}{3}} \\\\ y &=& \dfrac{ 2}{\dfrac{2}{3}} \\\\ \mathbf{y} &=& \mathbf{3} \\ \hline \end{array}\)

a.
The minimum value of the modulus of n is
\(\begin{array}{|rcll|} \hline && \mathbf{\sqrt{x^2+y^2}} \\ &=& \sqrt{1^2+3^2} \\ &=& \mathbf{\sqrt{10}} \\ \hline \end{array}\)

b.
One example of the complex number n is \(\mathbf{ 1+3i }\)

What is the complex number  z=−3+3√3i represented in polar form?

It's very difficult to read.

Jelly beans,

Please just ask one question per post.

Given a=5 and b=12,

c = 13

∠α = 22.62° = 22°37'12" = 0.39479 rad

∠β = 67.38° = 67°22'48" = 1.17601 rad

area = 30

perimeter = 30

https://www.calculator.net/pythagorean-theorem-calculator.html?av=5&asv=&bv=12&bsv=&cv=&csv=&x=46&y=23

This is real math! 

The area of the square is       A = 256 u²  

As follows:

The area of the square is  324 u²  

I made a boo-boo         it's not 324 u² 

Let M, N, and P be the midpoints of sides\( \overline{ZX}\), \(\overline{XY}\), and \(\overline{YZ}\) of triangle YZX, respectively.
Let \(\overline{XT}\) be an altitude of the triangle.
If \(\angle ZYX = 71^\circ\), \(\angle YZX = 36^\circ\), and \(\angle ZXY = 73^\circ\), then what is \(\angle NTM + \angle NPM\) in degrees?

\(\text{we have $\overline{XY} \parallel \overline{MP} $ and $\overline{XZ} \parallel \overline{NP}$} \\ \text{so $\angle YZX = \angle NPY$ and $\angle ZYX = \angle MPZ$ }\)

\(\begin{array}{|rcll|} \hline \mathbf{\angle NPM} = \ ? \\ \hline \angle NPM &=& 180^\circ - (36^\circ+71^\circ) \\ \angle NPM &=& 180^\circ - 107^\circ \\ \mathbf{\angle NPM} &=& \mathbf{73^\circ} \\ \hline \end{array}\)

\(\text{$\triangle NYT$ is isosceles, so $\angle NYT = \mathbf{\angle YTN = 71^\circ}$ }\)

\(\text{$\triangle MTZ$ is isosceles, so $\angle TZM = \mathbf{\angle MTZ = 36^\circ}$ }\)

\(\begin{array}{|rcll|} \hline \angle NTM + \angle NPM &=& 73^\circ+73^\circ \\ \mathbf{\angle NTM + \angle NPM} &=& \mathbf{146^\circ} \\ \hline \end{array}\)

In rectangle ABCD, AB = 3 and BC = 5.  The corner at C is folded up and lands on AB so that AC = 2 and BC = 1.  Find the area of quadrilateral CPNM.

The area of quadrilateral CPNM  is:                A ≈ 6.067 u²  

The area of the square is  324 u²   

What is the component form of resultant of \(4\vec{b} -5\vec{a}\) ?

\(\vec{a} ~=~ <1,-3> \\ \vec{b} ~=~ <-1,2> \\ 4\vec{b} -5\vec{a} ~=~ < ~, ~> \)

\(\begin{array}{|rcll|} \hline \mathbf{4\vec{b} -5\vec{a}} &=& 4\dbinom{-1}{2} -5\dbinom{1}{-3} \\\\ &=& \dbinom{4(-1)-5(1)}{4\cdot 2-5(-3)} \\\\ &=& \dbinom{-4-5}{8+15} \\\\ &=& \dbinom{-9}{23} \\\\ \mathbf{4\vec{b} -5\vec{a}} &=& \mathbf{<-9,~23>} \\ \hline \end{array}\)

Simplify

\(\sqrt{1 + \dfrac{\sqrt{3}}{2}} + \sqrt{1 - \dfrac{\sqrt{3}}{2}}\)

\(\begin{array}{|rcll|} \hline && \mathbf{ \sqrt{1 + \dfrac{\sqrt{3}}{2}} + \sqrt{1 - \dfrac{\sqrt{3}}{2}} } \\\\ &=& \sqrt{ \left(\sqrt{1 + \dfrac{\sqrt{3}}{2}} + \sqrt{1 - \dfrac{\sqrt{3}}{2}}\right)^2 } \\\\ &=& \sqrt{ \left(\sqrt{1 + \dfrac{\sqrt{3}}{2}}\right)^2 +2 \sqrt{1 + \dfrac{\sqrt{3}}{2}}\sqrt{1 - \dfrac{\sqrt{3}}{2}} + \left(\sqrt{1 - \dfrac{\sqrt{3}}{2}}\right)^2 } \\\\ &=& \sqrt{ 1 + \dfrac{\sqrt{3}}{2} + 2\sqrt{ \left(1 + \dfrac{\sqrt{3}}{2}\right) \left(1 - \dfrac{\sqrt{3}}{2}\right) } + 1 - \dfrac{\sqrt{3}}{2} } \\\\ &=& \sqrt{ 2 + 2\sqrt{ \left(1 + \dfrac{\sqrt{3}}{2}\right) \left(1 - \dfrac{\sqrt{3}}{2}\right) } } \\\\ &=& \sqrt{ 2 + 2\sqrt{ 1^2-\left( \dfrac{\sqrt{3}}{2} \right)^2 } } \\\\ &=& \sqrt{ 2 + 2\sqrt{ 1 - \dfrac{3}{4} } } \\\\ &=& \sqrt{ 2 + 2\sqrt{ \dfrac{1}{4} } } \\\\ &=& \sqrt{ 2 + 2\cdot \dfrac{1}{2} } \\\\ &=& \sqrt{ 2 + 1 } \\\\ &=& \mathbf{\sqrt{3}} \\ \hline \end{array}\)