heureka

a 6 sided shape has a perimeter of 62 inches one side is 11 inches long and the other side is 11 inches long

the lengths of the other 4 sides are equal what is one of those 4 sides.

Let x = the other 4 sides

\(\begin{array}{|rcll|} \hline 62 &=& 11 + 11 + 4x \\ 62 &=& 22 + 4x \\ 62-22 &=& 4x \\ 40 &=& 4x \quad & | \quad : 4 \\ 10 &=& x \\ \hline \end{array}\)

The other 4 sides are 10 inches long.

A market had only cookies and donuts.

The donuts were the 3/5 of the market and were 5 more than cookies.

How many cookies were there?

Let d = donuts
Let c = cookies

\(\begin{array}{|lrcll|} \hline (1) & c &=& d-5 \\\\ (2) & \frac{3}{5}\cdot ( c+d) &=& d \quad & | \quad c=d-5 \\ & \frac{3}{5}\cdot ( d-5+d) &=& d \\ & \frac{3}{5}\cdot ( 2d-5 ) &=& d \\ & 3\cdot ( 2d-5 ) &=& 5d \\ & 6d-15 &=& 5d \\ & \mathbf{d} & \mathbf{=} & \mathbf{15} \\\\ & c &=& d-5 \\ & c &=& 15-5 \\ & \mathbf{c} & \mathbf{=} & \mathbf{10} \\ \hline \end{array}\)

There were 10 cookies.

how many times do you have to divide-243 by 3 to get a quotient of -1?

\(\begin{array}{|rcll|} \hline \dfrac{-243}{3} &=& -81 \\\\ \dfrac{-81}{3} &=& -27 \\\\ \dfrac{-27}{3} &=& -9 \\\\ \dfrac{-9}{3} &=& -3 \\\\ \dfrac{-3}{3} &=& -1 \\ \hline \end{array} \)

5 times.

how do i plug in the unit vector?

Each unit vector has the length 1

\(\text{2D unit vector } \vec{v}_0 = \frac{\vec{v}}{v} \qquad \text{ with } \qquad \vec{v} = \binom {v_x}{v_y} \qquad \text{ and} \qquad v = \sqrt{v_x^2+v_y^2} \)

\(\text{3D unit vector } \vec{v}_0 = \frac{\vec{v}}{v} \qquad \text{ with } \qquad \vec{v} = \begin{pmatrix} v_x \\ v_y \\ v_z \end{pmatrix} \qquad \text{ and} \qquad v = \sqrt{v_x^2+v_y^2+v_z^2} \)

If a reflection in the line y = -x occurs, then the rule for this reflection is:

a (x, y) in (y, x)

b (x, y) in (-y, -x)

c (x, y) in (x, -y)

d (x, y) in (-x, y)

The rule for this reflection is

b (x, y) in (-y, -x)

In the figure, ABCD and BEFG are squares, and BCE  is an equilateral triangle.

What is the number of degrees in angle GCE ?

Center of a circle in B with radius r=BG=BE=BC

Central Angle Theorem:

The angle at the centre is twice the angle at the circumference.

\(\beta = \frac{\alpha}{2}=\frac{90^{\circ}}{2}=45^{\circ}\)

What is the equation of the parabola passing through (1,5), (0,6), and (2,3)?

Formula parabola:

\(\begin{array}{|rcll|} \hline y = ax^2+bx+c \\ \hline \end{array} \)

a, b, c = ?

\(\begin{array}{|lrcll|} \hline P(0,6): & 6 &=& 0^2\cdot a+0\cdot b+c \\ & 6 &=& c \\\\ P(1,5): & 5 &=& 1^2\cdot a+1\cdot b+c \\ & 5 &=& a+b+c \quad & | \quad c=6 \\ (1) & 5 &=& a+b+6 \\\\ P(2,3): & 3 &=& 2^2\cdot a+2\cdot b+c \\ & 3 &=& 4a+2b+c \quad & | \quad c=6 \\ (2) & 3 &=& 4a+2b+6 \\ \hline \end{array} \)

a, b = ?

\(\begin{array}{|rcll|} \hline (2) & 3 &=& 4a+2b+6 \quad & | \quad : 2\\ & 1.5 &=& 2a+b+3 \\\\ (1) & 5 &=& a+b+6 \\ \hline (2)-(1): & 1.5-5 &=& 2a+b+3- (a+b+6) \\ & -3.5 &=& 2a+b+3- a-b-6 \\ & -3.5 &=& a-3 \\ & -0.5 &=& a \\\\ & 5 &=& a+b+6 \quad & | \quad a=-0.5 \\ & 5 &=& -0.5+b+6 \\ & 5 &=& 5.5+b \\ & 5-5.5 &=& b \\ & -0.5 &=& b \\ \hline \end{array} \)

Formula parabola:

\(\begin{array}{|rcll|} \hline y &=& ax^2+bx+c \quad & | \quad a=-0.5 \quad b=-0.5 \quad c=6 \\ \mathbf{y} & \mathbf{=} & \mathbf{-0.5x^2-0.5x+6} \\ \hline \end{array}\)

Hey, I'm studying for a test on Wednesday.

Would I be able to cancel out the "^16" in this equation to get lim_(x->∞) (x + 1)^16/x^16 = 1?

\(\begin{array}{|rcll|} \hline && \lim \limits_{x\to \infty} \dfrac{ (x + 1)^{16} } {x^{16}} \\ &=& \lim \limits_{x\to \infty} \dfrac{ \binom{16}{0} x^{16} + \binom{16}{1} x^{15} + \binom{16}{2} x^{14} + \ldots + \binom{16}{15} x^1 + \binom{16}{16} } {x^{16}} \\ &=& \lim \limits_{x\to \infty} \dfrac{ 1\cdot x^{16} + 16 \cdot x^{15} + 120 \cdot x^{14} + \ldots + 16\cdot x + 1 } {x^{16}} \\ &=& \lim \limits_{x\to \infty} \frac{ x^{16} } {x^{16}} + 16 \cdot \frac{ \cdot x^{15}} {x^{16}} + 120 \cdot \frac{ x^{14}} {x^{16}} + \ldots + 16\cdot \frac{ x } {x^{16}} + \frac{ 1 } {x^{16}} \\ &=& \lim \limits_{x\to \infty} 1 + \frac{ 16} {x} + \frac{ 120} {x^2} + \ldots + \frac{ 16 } {x^{15}} + \frac{ 1 } {x^{16}} \\ &=& 1 + 0 + 0 + \ldots + 0 + 0 \\ &=& 1 \\ \hline \end{array} \)

simplify sin (π+x) - sin (π-x)

Because \(\sin(\pi+x) = - \sin(\pi-x)\)

\(\begin{array}{|rcll|} \hline && \sin(\pi+x) - \sin(\pi-x) \\ &=& -\sin(\pi-x) - \sin(\pi-x) \\ &=& -2\cdot \sin(\pi-x) \quad & | \quad \sin(\pi-x) = \sin(x)\\ &=& -2\cdot \sin(x) \\ \hline \end{array}\)

Welchen einmaligen Betrag muss ein Lotteriegewinner in ein Sparkonto mit einem Zinssatz von 3,9 % p.a. einzahlen, um anschließend 25 Jahre lang jeweils am Jahresende 48000€ davon abheben zu können?

Wieviel € sollte die Anlagesumme betragen?(auf den Cent genau)

What amount does a lottery winner have to pay into a savings account with an interest rate of 3.9% pa and then be able to withdraw 48,000 € for 25 years at the end of the year?

How much € should the amount be? (To the cent exactly)

Sei C = 48000 €

Sei der Zinssatz = i = 3.9 %

Sei die Anzahl der Jahre = n

Sei das Anfangskapital = K

Das Kapital nach dem 1. Jahr beträgt:

\(\begin{array}{|rcll|} \hline K \cdot(1+i) - C \\ \hline \end{array}\)

Das Kapital nach dem 2. Jahr beträgt:

\(\begin{array}{|rcll|} \hline && [K \cdot(1+i) - C]\cdot (1+i) -C \\ &=& K \cdot(1+i)^2 - C\cdot (1+i) -C \\ &=& K \cdot(1+i)^2 - C\cdot [1+(1+i)] \\ \hline \end{array}\)

Das Kapital nach dem 3. Jahr beträgt:

\(\begin{array}{|rcll|} \hline && \Big( K \cdot(1+i)^2 - C\cdot [1+(1+i)]\Big) \cdot (1+i) -C \\ &=& K \cdot(1+i)^3 - C\cdot [1+(1+i)+(1+i)^2] \\ \hline \end{array} \)

Das Kapital nach dem n. Jahr beträgt:

\(\begin{array}{|rcll|} \hline &=& K \cdot(1+i)^n - C\cdot [ \underbrace{ 1+(1+i)+(1+i)^2+\ldots + (1+i)^{n-1} }_{=\frac{(1+i)^n-1}{i} } ] \\ &=& K \cdot(1+i)^n - C\cdot \Big(\frac{(1+i)^n-1}{i} \Big) \\ \hline \end{array} \)

Das Kapital beträgt am Ende 0 €:

\(\small{ \begin{array}{|rcll|} \hline 0\ €&=& K \cdot(1+i)^n - C\cdot \Big(\frac{(1+i)^n-1}{i} \Big) \\ K \cdot(1+i)^n &=& C\cdot \Big(\frac{(1+i)^n-1}{i} \Big) \\ K &=& C\cdot \Big(\frac{(1+i)^n-1}{i\cdot (1+i)^n} \Big) \\ K &=& C\cdot \Big( \frac{\frac{(1+i)^n}{(1+i)^n} -\frac{1}{(1+i)^n} }{i} \Big) \\ K &=& C\cdot \Big( \dfrac{1 - (1+i)^{-n} }{i} \Big) \qquad \text{This is exact the Formula from Melody - Dies ist genau die Formel von Melody } \\ \hline \end{array} } \)

Wir setzen ein:

\(\begin{array}{|rcll|} \hline K &=& C\cdot \Big( \dfrac{1 - (1+i)^{-n} }{i} \Big) \quad & | \quad i = 3.9 \% \quad n=25\ \text{Jahre} \quad C=48000\ € \\ &=& 48000\cdot \Big( \dfrac{1 - (1+3.9 \%)^{-25} }{3.9 \%} \Big) \\ &=& 48000\cdot \Big( \dfrac{1 - 1.039^{-25} }{0.039} \Big) \\ &=& 48000\cdot \Big( \dfrac{1 - 0.38424773049 }{0.039} \Big) \\ &=& 48000\cdot \Big( \dfrac{0.61575226951 }{0.039} \Big) \\ &=& 48000\cdot 15.7885197311 \\ &=& 757848.947094\ € \\ \hline \end{array} \)

Der Lotteriegewinner muss 757 848.95 € in ein Sparkonto einzahlen.

The lottery winner have to pay into a savings account about 757 848.95 €