heureka

Lösungen:

Find the Cartesian equivalent to the parametric equation x(t)= 3-2t and y(t)=(t^2) -2t

$\begin{array}{|rcll|} \hline x(t)&=& 3-2t \\ x-3 &=& 2t \\ \mathbf{2t} &\mathbf{=}& \mathbf{ x-3 }\\ \mathbf{t} &\mathbf{=}& \mathbf{\frac12 (x-3) } \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline y(t)&=&(t^2) -2t \qquad &| \qquad \mathbf{t =\frac12 (x-3) } \\ y &=&[ \frac12 (x-3) ]^2 -(2t) \qquad &| \qquad \mathbf{2t = x-3 }\\ y &=&[ \frac12 (x-3) ]^2 -(x-3) \\ y &=&\frac14 (x-3)^2 -(x-3) \\ y &=&\frac14 (x-3)^2 -(x-3)\cdot \frac44 \\ y &=&\frac14 (x-3)\cdot [(x-3)-4] \\ y &=&\frac14 (x-3)\cdot [ x-3 -4] \\ y &=&\frac14 (x-3)\cdot ( x-7 ) \\ y &=&\frac14 (x^2-7x-3x+21) \\ \mathbf{y} &\mathbf{=}&\mathbf{ \frac14 \cdot (x^2-10x+21) }\\ \hline \end{array}$

2. Find the smallest positive integer that satisfies the system of congruences $\begin{align*} N &\equiv 2 \pmod{11}, \\ N &\equiv 3 \pmod{17}. \end{align*}$

see (a) in

What is the ones digit for 3^1555

$\begin{array}{|rcll|} \hline && 3^{1555} \pmod{10} \\ &\equiv& 3^{4\cdot 388 + 3} \pmod{10} \\ &\equiv& 3^{4\cdot 388}\cdot 3^3 \pmod{10} \\ &\equiv& (3^{4})^{388} \cdot 3^3 \pmod{10} \qquad | \qquad 3^4 \equiv 1 \pmod{10} \\ &\equiv& (1)^{388} \cdot 3^3 \pmod{10} \\ &\equiv& 1 \cdot 3^3 \pmod{10} \\ &\equiv& 3^3 \pmod{10} \\ &\equiv& 27 \pmod{10} \\ &\equiv& 7 \pmod{10} \\ \hline \end{array}$

A=3i+2j-6k ,

B=2i+4j+7k and

C=2i+9j+3k

the magnitude of vector D=A+B-C is?

$\begin{array}{|rcll|} \hline D &=& A+B-C \\ D &=& (3i+2j-6k)+ (2i+4j+7k)- (2i+9j+3k) \\ D &=& (3+2-2)i+(2+4-9)j+(-6+7-3)k \\ D &=& 3i-3j-2k \\ \left\lVert D \right\rVert &=& \sqrt{3^2+(-3^2)+(-2)^2} \\ \left\lVert D \right\rVert &=& \sqrt{9+9+4} \\ \left\lVert D \right\rVert &=& \sqrt{22} \\ \hline \end{array}$

The magnitude of vector D=A+B-C is $\sqrt{22} = 4.69041575982$

|3k-2|=2|k+2|

Formula:

$\begin{array}{|rcll|} \hline |a|^2 &=& a^2 \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline |3k-2| &=& 2|k+2| \qquad | \qquad \text{square both sides}\\ (3k-2)^2 &=& 2^2\cdot (k+2)^2 \\ 9k^2-12k+4 &=& 4\cdot (k^2+4k+4) \\ 9k^2-12k+4 &=& 4k^2+16k+16 \\ \dots \\ 5k^2-28k-12 &=& 0 \\\\ k_{1,2} &=& \frac{28 \pm \sqrt{28^2-4\cdot 5\cdot (-12) } }{2\cdot 5} \\ k_{1,2} &=& \frac{28 \pm \sqrt{1024 } }{10} \\ k_{1,2} &=& \frac{28 \pm 32 } {10} \\\\ k_1 &=& \frac{28 + 32 } {10} \\ k_1 &=& \frac{ 60 } {10} \\ \mathbf{ k_1 }& \mathbf{=} & \mathbf{6} \\\\ k_2 &=& \frac{28 - 32 } {10} \\ k_2 &=& \frac{ -4 } {10} \\ \mathbf{ k_2 } & \mathbf{=} & \mathbf{-0.4} \\ \hline \end{array}$

fully simplify (4a-b)^4

$\begin{array}{|rcll|} \hline && \mathbf{ (4a-b)^4 }\\ &=& (4a)^4 - 4\cdot(4a)^3\cdot b^1 + 6\cdot (4a)^2\cdot b^2 - 4\cdot (4a)^1\cdot b^3 + b^4 \\ &=& 4^4a^4 - 4\cdot4^3a^3\cdot b + 6\cdot 4^2a^2\cdot b^2 - 4\cdot 4a \cdot b^3 + b^4 \\ &=& 256a^4 - 4^4a^3\cdot b + 6\cdot 16a^2\cdot b^2 - 16a \cdot b^3 + b^4 \\ &\mathbf{=}& \mathbf{256 a^4 -256 a^3 b+96 a^2 b^2-16 a b^3+b^4 } \\ \hline \end{array}$

Positive integer x leaves a remainder of 2 when divided by 8.
What is the remainder when x+9 is divided by 8

$\begin{array}{|rcll|} \hline x & \equiv & 2 \pmod{8} \\ x+{\color{red} 9 } & \equiv & 2+{\color{red} 9 } \pmod{8} \\ x+9 & \equiv & 11 \pmod{8} \\ x+9 & \equiv & 11-8 \pmod{8} \\ x+9 & \equiv & 3 \pmod{8} \\ \hline \end{array}$

If the terminal side of 'theta' is in quadrant II and the reference angle is pi/3,

find the exact values of sin(theta) and cos(theta)

II. Quadrant : $\cos(\theta) <0 \\ \sin(\theta) >0$

So the answer is c.

$\begin{array}{|rcll|} \hline \sin^2(\theta) + \cos^2(\theta) &=& 1 \\ \left( \frac{\sqrt{3}}{2} \right)^2 + \left( -\frac12 \right)^2 &=& 1 \\ \frac{3}{4} + \frac14 &=& 1 \\ 1 &=& 1 \\ \hline \end{array}$

Find the M is the midpoint of the line segment PQ, what is the coordinates of point Q? P (4, 5), M (-2, -3)

Answers:

$a. (2,2) \\ b. (6,8) \\ c. (-8,-11) \\ d. (-8,-15) \\ e. (-4, -7.5) \\$

$\begin{array}{|rcll|} \hline \frac{\vec{P} + \vec{Q}} {2} &=& \vec{M} \quad & | \quad \cdot 2\\\\ \vec{P} + \vec{Q} &=& 2\cdot \vec{M} \quad & | \quad - \vec{P} \\\\ \vec{Q} &=& 2\cdot \vec{M} - \vec{P} \quad & | \quad \vec{M} = \binom{-2}{-3} \quad \vec{P} = \binom{4}{5}\\\\ \vec{Q} &=& 2\cdot \binom{-2}{-3} - \binom{4}{5} \\\\ \vec{Q} &=& \binom{-2\cdot 2}{-3\cdot 2} - \binom{4}{5} \\\\ \vec{Q} &=& \binom{-4}{-6} - \binom{4}{5} \\\\ \vec{Q} &=& \binom{-4-4}{-6-5} \\\\ \mathbf{ \vec{Q} } &\mathbf{=} & \mathbf{ \binom{-8}{-11} } \\\\ \hline \end{array}$

Point Q is  (-8, -11)

(c.)