heureka

If

\(a + b + c = 0 \)

and

\(a^3 + b^3 + c^3 = 216\),

find the value of \(abc\).

Formula:  \(\boxed{a^3+b^3+c^3 = 3abc+(a+b+c)(a^2+b^2+c^3-ab-bc-ca)}\)

\(\begin{array}{|rcll|} \hline \underbrace{a^3+b^3+c^3}_{=216} &=& 3abc+\underbrace{(a+b+c)}_{=0}(a^2+b^2+c^3-ab-bc-ca) \\\\ 216 &=& 3abc+0 \\ 3abc &=& 216 \quad | \quad : 3 \\ \mathbf{abc} &=&\mathbf{72} \\ \hline \end{array}\)

Find the length of the line segments whose endpoints have polar coordinate (7, 40 degres) and (15, 100 degrees).

\(\text{Let $P_1 =\dbinom{x_1}{y_1}=\dbinom{7\cos(40^\circ)}{7\sin(40^\circ) } $} \\ \text{Let $P_2 =\dbinom{x_2}{y_2}=\dbinom{15\cos(100^\circ)}{15\sin(100^\circ)} $} \)

\(\begin{array}{|rcll|} \hline \mathbf{ \overline{P_1P_2} } &=& \mathbf{ \sqrt{ \left( x_2-x_1\right)^2 + \left(y_2-y_1 \right)^2 } } \\\\ &=& \sqrt{ \Big( 15\cos(100^\circ) - 7\cos(40^\circ)\Big)^2 + \Big( 15\sin(100^\circ) - 7\sin(40^\circ)\Big)^2 } \\ &=& \tiny{\sqrt{ 15^2\cos^2(100^\circ) -2*15*7\cos(100^\circ)\cos(40^\circ) + 7^2\cos^2(40^\circ) + 15^2\sin^2(100^\circ) -2*15*7\sin(100^\circ)\sin(40^\circ) + 7^2\sin^2(40^\circ) }} \\ &=& \tiny{\sqrt{ 7^2\Big(\underbrace{\sin^2(40^\circ)+\cos^2(40^\circ)}_{\small{=1}}\Big) + 15^2\Big(\underbrace{\sin^2(100^\circ)+\cos^2(100^\circ)}_{\small{=1}}\Big) -2*15*7\Big(\underbrace{\cos(100^\circ)\cos(40^\circ) + \sin(100^\circ)\sin(40^\circ)}_{\small{=\cos(100^\circ-40^\circ)}} \Big) }} \\ &=& \sqrt{ 7^2+ 15^2 -2*15*7\cos(100^\circ-40^\circ) } \\ &=& \sqrt{ 7^2+ 15^2 -2*15*7\cos(60^\circ) } \quad | \quad \cos(60^\circ)= \dfrac{1}{2} \\ &=& \sqrt{ 7^2+ 15^2 - 15*7 } \\ &=& \sqrt{ 169 } \\ &=& \mathbf{13} \\ \hline \end{array}\)

The length of the line is 13

2.

Find the remainder when r^14 is divided by r - 1.

The remainder is 1

1.
When \(x^4 + 6x^3 - ax^2 - 45x -15\) is divided by \(x^2 - x - 6\) the remainder is \(3\).
What is \(a\)?

\(\text{Let $x^2 - x - 6 = (x-3)(x+2)$}\)

\(\begin{array}{|lrcll|} \hline & \mathbf{\dfrac{x^4 + 6x^3 - ax^2 - 45x -15}{(x-3)(x+2)}} &=& \mathbf{q(x) + \dfrac{3}{(x-3)(x+2)}} \quad | \quad \times (x-3)(x+2) \\\\ x=3: & x^4 + 6x^3 - ax^2 - 45x -15 &=& q(x)(x-3)(x+2) + 3 \\ & 3^4+6(3^3)-a(3^2)-45(3)-15 &=& q(x)(0)(5) + 3 \\ & 3^4+6(3^3)-a(3^2)-45(3)-15 &=& 3 \\ & 81+162-9a-135-15 &=& 3 \\ & 93-9a &=& 3 \quad | \quad :3 \\ & 31-3a &=& 1 \\ & 3a &=& 31-1 \\ & 3a &=& 30 \quad | \quad :3 \\ & \mathbf{a} &=& \mathbf{10} \\ \hline \end{array}\)

Thank you, Roxettna !

A surveyor determines that the angle of elevation to the top of a vertical building from a pt on level ground is 38deg.

She then moves back 50 feet & determines that the angle of elevation is 32 deg.

Find the height of the building.

\(\begin{array}{|rcll|} \hline \dfrac{\sin(6^\circ)}{50} &=& \dfrac{\sin(32^\circ)}{y} \\ \mathbf{y} &=& \mathbf{ \dfrac{50\sin(32^\circ)}{\sin(6^\circ)} } \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \sin(38^\circ) &=& \dfrac{h}{y} \\ h &=& y\sin(38^\circ) \\ h &=& \dfrac{50\sin(32^\circ)\sin(38^\circ)}{\sin(6^\circ)} \\ \mathbf{h} &=& \mathbf{ 156\ \text{feet} } \\ \hline \end{array} \)

Find

\(\sin\left( \arcsin\left(\dfrac{3}{5}\right) + \arccos\left(\dfrac{15}{17}\right) \right)\).

\(\text{Let $x=\dfrac{3}{5},\ y=\dfrac{15}{17} $}\)

\(\begin{array}{|rcll|} \hline && \mathbf{\sin\left( \arcsin\left(\dfrac{3}{5}\right) + \arccos\left(\dfrac{15}{17}\right) \right)} \\\\ &=& \sin\Big( \underbrace{\arcsin\left(x\right)}_{\alpha} + \underbrace{\arccos\left(y\right)}_{\beta} \Big) \\ && \boxed{\text{Formula: } \sin(\alpha+\beta)=\sin(\alpha)\cos(\beta)+ \cos(\alpha)\sin(\beta)} \\\\ &=& \sin\Big(\arcsin\left(x\right)\Big)\cos\Big(\arccos\left(y\right)\Big) + \cos\Big(\arcsin\left(x\right)\Big)\sin\Big(\arccos\left(y\right)\Big) \\ &=& xy + \mathbf{\cos\Big(\arcsin\left(x\right)\Big)}\sin\Big(\arccos\left(y\right)\Big) \\ && \boxed{ \arcsin\left(x\right) = \alpha\\ x=\sin(\alpha)\\ x^2=\sin^2(\alpha)\\ x^2=1-\cos^2(\alpha)\\ \cos^2(\alpha)=1-x^2 \\ \cos(\alpha) =\sqrt{1-x^2} \\ \mathbf{\cos\Big(\arcsin\left(x\right)\Big) =\sqrt{1-x^2} } } \\\\ &=& xy + \sqrt{1-x^2}\sin\Big(\arccos\left(y\right)\Big) \\ && \boxed{ \arccos\left(y\right) = \beta\\ y=\cos(\beta)\\ y^2=\cos^2(\beta)\\ y^2=1-\sin^2(\beta)\\ \sin^2(\beta)=1-y^2 \\ \sin(\beta) =\sqrt{1-y^2} \\ \mathbf{\sin\Big(\arccos\left(y\right)\Big) =\sqrt{1-y^2} } } \\\\ &=& xy + \sqrt{1-x^2}\sqrt{1-y^2} \quad | \quad x=\dfrac{3}{5},\ y=\dfrac{15}{17} \\ &=& \dfrac{3}{5}*\dfrac{15}{17} + \sqrt{1-\dfrac{3^2}{5^2}}\sqrt{1-\dfrac{15^2}{17^2}} \\ &=& \dfrac{45}{85} + \dfrac{4}{5} * \dfrac{8}{17} \\ &=& \dfrac{45}{85} + \dfrac{32}{85} \\ &=& \mathbf{ \dfrac{77}{85} } \\ \hline \end{array}\)

For
x > 1, simplify \(\sqrt{ x + \sqrt{2x - 1} } - \sqrt{ x - \sqrt{2x - 1} }\).

\(\text{Let $k= \sqrt{2x - 1},$ or $k^2=2x-1$}\)

\(\begin{array}{|rcll|} \hline && \mathbf{ \sqrt{ x + \sqrt{2x - 1} } - \sqrt{ x - \sqrt{2x - 1} } } \\\\ &=& \sqrt{ x + k } - \sqrt{ x - k } \\\\ &=& \sqrt{ \left( \sqrt{ x + k } - \sqrt{ x - k } \right)^2 } \\\\ &=& \sqrt{ (x+k) -2\sqrt{ (x+k)(x-k) } + (x-k) } \\\\ &=& \sqrt{ 2x -2\sqrt{ x^2-k^2 } } \quad | \quad k^2=2x-1 \\\\ &=& \sqrt{ 2x -2\sqrt{ x^2-(2x-1) } } \\\\ &=& \sqrt{ 2x -2\sqrt{ x^2-2x+1 } } \\\\ &=& \sqrt{ 2x -2\sqrt{ (x-1)^2 } } \\\\ &=& \sqrt{ 2x -2 (x-1) } \\\\ &=& \sqrt{ 2x -2x +2 } \\\\ &=& \mathbf{ \sqrt{ 2 } } \quad | \quad x \geq 1 \\ \hline \end{array}\)

Is it possible cut the obtuse triangle into acute triangular pieces?

Part 2

(b) Find the sum of all the numbers on the black squares. (|a| < 1 and |b| < 1)

Sum of all the numbers on the black squares:

\(1 \\ +a^2+ab+b^2 \\ +a^4+a^3b+a^2b^2+ab^3+b^4 \\ +a^6+a^5b+a^4b^2+a^3b^3+a^2b^4+ab^5+b^6 \\ +\ldots\)

Infinite sum:
\(\sum \limits_{l=0}^{\infty} \left( \sum \limits_{k=0}^{2l} a^{2l-k}b^k \right) \)

\(\begin{array}{|rcll|} \hline && \mathbf{\sum \limits_{l=0}^{\infty} \left( \sum \limits_{k=0}^{2l} a^{2l-k}b^k \right)} \\ &=& \sum \limits_{l=0}^{\infty} \left( \sum \limits_{k=0}^{2l} a^{2l}a^{-k}b^k \right) \\ &=& \sum \limits_{l=0}^{\infty} a^{2l}\left( \sum \limits_{k=0}^{2l} a^{-k}b^k \right) \\ &=& \sum \limits_{l=0}^{\infty} a^{2l} \left( \sum \limits_{k=0}^{2l} \left(\dfrac{b}{a}\right)^k \right) \quad | \quad \sum \limits_{k=0}^{2l} \left(\dfrac{b}{a}\right)^k = \dfrac{1- \left(\dfrac{b}{a}\right)^{2l+1} }{1-\dfrac{b}{a} } \\ &=& \sum \limits_{l=0}^{\infty} a^{2l} \left( \dfrac{1- \left(\dfrac{b}{a}\right)^{2l+1} }{1-\dfrac{b}{a} } \right) \\ &=& \sum \limits_{l=0}^{\infty} a^{2l} \left( \dfrac{a- \left(\dfrac{b}{a}\right)^{2l}b }{a-b } \right) \\ &=& \dfrac{1}{(a-b)} \sum \limits_{l=0}^{\infty} a^{2l} \left( a- \left(\dfrac{b}{a}\right)^{2l}b \right) \\ &=& \dfrac{1}{(a-b)} \sum \limits_{l=0}^{\infty} \left( a^{2l+1}- \left(\dfrac{ a^{2l}b^{2l}}{a^{2l}}\right)b \right) \\ &=& \dfrac{1}{(a-b)} \sum \limits_{l=0}^{\infty} \left( a^{2l+1}- b^{2l+1} \right) \\ &=& \dfrac{1}{(a-b)} \left( \sum \limits_{l=0}^{\infty} \left( a^{2l+1} \right) - \sum \limits_{l=0}^{\infty} \left( b^{2l+1} \right) \right) \\ &=& \dfrac{1}{(a-b)} \left( a\sum \limits_{l=0}^{\infty} a^{2l} - b\sum \limits_{l=0}^{\infty} b^{2l} \right) \quad | \quad \sum \limits_{l=0}^{\infty} a^{2l} = \dfrac{1}{1-a^2},\ \sum \limits_{l=0}^{\infty} b^{2l} = \dfrac{1}{1-b^2} \\ &=& \dfrac{1}{(a-b)} \left( \dfrac{a}{1-a^2} - \dfrac{b}{1-b^2} \right) \\ &=& \dfrac{1}{(a-b)} \left( \dfrac{a(1-b^2)-b(1-a^2)}{(1-a^2)(1-b^2)} \right) \\ &=& \dfrac{1}{(a-b)} \left( \dfrac{a-ab^2-b+ba^2}{(1-a^2)(1-b^2)} \right) \\ &=& \dfrac{1}{(a-b)} \left( \dfrac{(a-b)+ab(a-b)}{(1-a^2)(1-b^2)} \right) \\ &=& \dfrac{1}{(a-b)} \left( \dfrac{(a-b)(1+ab)}{(1-a^2)(1-b^2)} \right) \\ &=& \mathbf{ \dfrac{1+ab}{(1-a^2)(1-b^2)} } \\ \hline \end{array}\)