heureka

tg15∘⋅tg25∘⋅tg35∘⋅tg85∘tg15∘⋅tg25∘⋅tg35∘⋅tg85∘=?

$\tan(15°)\cdot \tan(25°)\cdot \tan(35°)\cdot \tan(85°)=1$

$\tan(15°)\cdot \tan(25°)\cdot \tan(35°)\cdot \tan(85°)\times \tan(15°)\cdot \tan(25°)\cdot \tan(35°)\cdot \tan(85°) =1\times 1 = 1$

tg15∘⋅tg25∘⋅tg35∘⋅tg85=? help pls

$\tan(15°)\cdot \tan(25°)\cdot \tan(35°)\cdot \tan(85°)=1$

A market gardener has to fertilize a triangular field with sides of lengths 90m, 45m, and 65m.

The fertilizer is to be spread so that 1 kg covers 10m squared. One bag of fertilizer has a mass of 9.1 kg.

How many bags of fertilizer will he need?

Let A = the area of the triangular field

Let a = 90 m
Let b = 45 m
Let c = 65 m

Heron:

$\begin{array}{|rcll|} \hline A &=& \sqrt{s(s-a)(s-b)(s-c)} \quad & | \quad s = \frac{a+b+c}{2}= \frac{90+45+65}{2}=100 \\ A &=& \sqrt{100\cdot(100-90)\cdot(100-45)\cdot(100-65)} \\ A &=& \sqrt{100\cdot(10)\cdot(55)\cdot(35)} \\ A &=& 10\cdot\sqrt{10\cdot 55\cdot 35} \\ A &=& 10\cdot\sqrt{19250} \\ A &=& 10\cdot 138.744369255 \\ A &=& 1387.44369255\ m^2 \\ \hline \end{array}$

The area of the triangular field is 1387.44 m²

Let x = bags of fertilizer

$\begin{array}{|rcll|} \hline x &=& A \cdot \frac{1\ kg}{10\ m^2} \cdot \frac{1\ \text{bag}}{9.1\ kg} \quad & | \quad A &=& 1387.44369255\ m^2 \\ x &=& 1387.44369255\ m^2 \cdot \frac{1\ kg}{10\ m^2} \cdot \frac{1\ \text{bag}}{9.1\ kg} \\ x &=& \frac{1387.44369255 }{10\cdot 9.1 } \ \text{bags} \\ x &=& \frac{1387.44369255 }{91} \ \text{bags} \\ x &=& 15.2466339841\ \text{bags} \\ \hline \end{array}$

We need $\approx$16 bags

Two ships leave a port at the same time. One sails at 17km/h on a bearing of 205 degrees.

The other sails at 21 km/h on a bearing of 243 degrees.

How far apart are the ships after 2 hours?

After 2 hours:

Let d = the distance

$\begin{array}{|rcll|} \hline d^2 &=& 42^2 + 34^2 - 2 \cdot 42 \cdot 34 \cdot \cos(243^{\circ}-205^{\circ} ) \\ d^2 &=& 2920 - 2856 \cdot \cos(38^{\circ} ) \\ d^2 &=& 2920 - 2856 \cdot 0.78801075361 \\ d^2 &=& 2920 - 2250.55871230 \\ d^2 &=& 669.441287699 \\ d &=& 25.8735634905\ km \\ \hline \end{array}$

The distance after 2 hours is 25.874 km

Two office towers are 30 m apart. From the 15th floor (which is 40 m above the ground) of the shorter tower, the angle of elevation to the top of the other tower is 70 degrees. 

Determine the angle of depression to the base of the taller tower from the 15th floor.

Determine the height of the taller tower. 

Let $\varphi$ the angle of depression

$\begin{array}{|rcll|} \hline \tan{\varphi} &=& \frac{40\ m}{30\ m} \\ \tan{\varphi} &=& \frac{4}{3} \\ \varphi &=& 53.1301023542^{\circ} \\ \hline \end{array}$

The angle of depression to the base of the taller tower from the 15th floor is $53.1301^{\circ}$

Let H = height of the taller tower

$\begin{array}{|rcll|} \hline \tan{70^{\circ}} &=& \frac{H-40\ m}{30\ m} \\ 30\cdot \tan{70^{\circ}} &=& H-40 \\ H &=& 40 + 30\cdot \tan{70^{\circ}} \\ H &=& 40 + 30\cdot 2.74747741945 \\ H &=& 40 + 82.4243225836 \\ H &=& 122.424322584\ m \\ \hline \end{array}$

The height of the taller tower is 122.42 m

=

       4*1    
    + 1*3
    + 8*4  
    + 82*5
    + 4*8
    + 9*10
    + 8*20
    + 15*50
    + 10*100

= 2481

840 als produkt von primfaktoren

840 = 2³ * 3 * 5 * 7

41105*(1+x/100)^5+(1+x+1/100)^5=57975.9? Nach x auflösen..

x-(sin(2x))/2=1.1

$x-(sin(2x))/2=1.1\\ x-\frac{ 2\sin(x)\cdot\cos(x)}{2} =1.1 \\ x-\sin(x)\cdot\cos(x) =1.1 \\$

Let

$a \equiv 1 \pmod{4}$.

Find the value of

$6a + 5 \pmod{4}$

Express your answer as a residue between 0 and the modulus.

$\begin{array}{|rcll|} \hline && 6a + 5 \pmod{4} \\ &\equiv& 6a + 5 -4 \pmod{4} \\ &\equiv& 6a + 1 \pmod{4} \quad & | \quad a \equiv 1 \pmod{4} \\ &\equiv& 6\cdot 1 + 1 \pmod{4} \\ &\equiv& 7 -1\cdot 4 \pmod{4} \\ &\equiv& 7 -4 \pmod{4} \\ &\equiv& 3 \pmod{4} \\ \hline \end{array}$

The answer is 3.