heureka

\(\begin{array}{rcll} log_{15}(5-n) &=& log_{15}(4n-5) \\ 5-n &=& 4n-5 \\ 5 &=& 5n-5 \\ 10 &=& 5n \\ n &=& \frac{10}{5} \\ n &=& 2 \end{array}\)

log5(-a)=log5(29)

\(log_5(-a)=log_5(29)\\ -a = 29 \\ a = -29\)

2 hoch 5.299.999  ?

log6(3*b)=log6(5-(2*b))

\(\begin{array}{rcll} log_6(3b) &=& log_6(5-2b) \\ 3b &=& 5-2b\\ 5b &=& 5\\ b &=& 1 \end{array} \)

0.2x + 0.3(x-27000)=800 Solve for x

\(\begin{array}{rcll} 0.2 \cdot x + 0.3\cdot (x-27000) &=& 800 \\ 0.2 \cdot x + 0.3\cdot x- 0.3\cdot 27000 &=& 800 \\ 0.5\cdot x- 0.3\cdot 27000 &=& 800 \\ 0.5\cdot x- 8100 &=& 800 \qquad & | \qquad + 8100 \\ 0.5\cdot x &=& 800 + 8100 \\ 0.5\cdot x &=& 8900 \\ 0.5\cdot x &=& 8900 \qquad & | \qquad \cdot 2 \\ x &=& 17800 \\ \end{array}\)

direction angle = \(\varphi\)

magnitude = m

\(\begin{array}{rcll} \vec{F}_1 &=& -100\cdot \binom{ \sin{(25^{\circ})} }{ \cos{(25^{\circ})} } \\ \vec{F}_1 &=& -200\cdot \binom{ \sin{(80^{\circ})} }{ \cos{(80^{\circ})} } \\ \vec{F} &=& \vec{F}_1+ \vec{F}_2 \\ \vec{F} &=& -100\cdot \dbinom{ \sin{(25^{\circ})} }{ \cos{(25^{\circ})} }-200\cdot \dbinom{ \sin{(80^{\circ})} }{ \cos{(80^{\circ})} }\\ \vec{F} &=& -\left[ \dbinom{ 100\cdot \sin{(25^{\circ})}+ 200\cdot \sin{(80^{\circ})} }{ 100\cdot \cos{(25^{\circ})} +200\cdot \cos{(80^{\circ})}} \right] \\ \vec{F} &=& -\left[ \dbinom{ 100\cdot 0.42261826174+ 200\cdot 0.98480775301 }{ 100\cdot 0.90630778704 +200\cdot 0.17364817767 } \right] \\ \vec{F} &=& -\left[ \dbinom{ 42.2618261741+ 196.961550602 }{ 90.6307787037 + 34.7296355334 } \right] \\ \vec{F} &=& -\left[ \dbinom{ 239.223376777 }{ 125.360414237 } \right] \\\\ \tan{ ( \varphi ) } &=& \frac{239.223376777}{125.360414237} \\ \varphi &=& 62.3440776108^{\circ}\\\\ m &=& \sqrt{239.223376777^2+ 125.360414237^2 }\\ m &=& 270.079724256\\\\ \vec{F} &=& -m \cdot \dbinom{ \sin{(\varphi)} } { \cos{(\varphi)} } \\ \vec{F} &=& -270.079724256 \cdot \dbinom{ \sin{(62.3440776108^{\circ})} }{ \cos{(62.3440776108^{\circ})} } \end{array}\)

The magnitude is 270 pounds.

The direction is N\(62.3^{\circ}\)E

volume of a cylinder that is 150 ft tall and a radius of 50 ft

Volume = V
Tall = h
Radius = r

\(\begin{array}{rcll} V &=& \pi \cdot r^2 \cdot h \\ V &=& \pi \cdot (50\ ft.)^2 \cdot 150\ ft. \\ V &=& \pi \cdot 2500\ ft.^2 \cdot 150\ ft. \\ V &=& 375000\ ft.^3 \\ \end{array}\)

∫45dx + 6∫12dx

\(\begin{array}{rcll} \int {54\ dx} + 6\cdot \int {12\ dx} &=& 54\cdot \int {dx} + 6\cdot12\cdot \int {dx} \\ &=& 54\cdot x + 6\cdot12\cdot x\\ &=& 54\cdot x + 72\cdot x\\ &=& 126\cdot x\\ \end{array}\)

1.

\(\begin{array}{rcll} 9 \cdot (x-3) &=& 10\cdot 18 \\ 9x - 27 &=& 180 \\ 9x &=& 180 + 27 \\ 9x &=& 207 \\ x &=& 23 \end{array}\)

2.

\(\begin{array}{lcll} x^2 &=& 9\cdot (9+7) \\ x^2 &=& 9\cdot 16 \\ x^2 &=& 144 \\ x &=& 12 \\ \end{array}\)

15cos30  = ?

\(15\cdot \cos{(30)} = 15 \cdot \frac{ \sqrt{3} } {2} \\ = 7.5 \cdot \sqrt{3}\\ = 7.5 \cdot 0.86602540378 \\ = 6.49519052838 \)