Received notice from the bank of a dishonored check from Jackson Group, $535.00, plus $25.00 fee;

Mathematics

1 answer:

iren2701 [21]3 years ago

5 0

Answer:

Step-by-step explanation:

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Read 2 more answers

Not sure f either of my answers would satisfy the question in the second picture

xz_007 [3.2K]

$\bf \begin{array}{llll}
tan(\alpha)=-\cfrac{80}{15}\qquad II\\\\
x=-15\\
y=8\\
r=17
\end{array} \qquad \qquad 
\begin{array}{llll}
cos(\beta)=\cfrac{5}{6}\qquad I\\\\
x=5\\
y=\sqrt{11}\\
r=6
\end{array}\\\\
-----------------------------\\\\
sin({{ \alpha}} + {{ \beta}})=sin({{ \alpha}})cos({{ \beta}}) + cos({{ \alpha}})sin({{ \beta}})
\\\\\\
sin({{ \alpha}} + {{ \beta}})=\cfrac{8}{17}\cdot \cfrac{5}{6}+\cfrac{-15}{17}\cdot \cfrac{\sqrt{11}}{6}\implies \cfrac{40-15\sqrt{11}}{102}$

$\bf cos({{ \alpha}} + {{ \beta}})= cos({{ \alpha}})cos({{ \beta}})- sin({{ \alpha}})sin({{ \beta}})
\\\\\\
\cfrac{-15}{17}\cdot \cfrac{5}{6}-\cfrac{8}{17}\cdot \cfrac{\sqrt{11}}{6}\implies \cfrac{-75-8\sqrt{11}}{102}$

$\bf tan({{ \alpha}} + {{ \beta}}) = \cfrac{tan({{ \alpha}})+ tan({{ \beta}})}{1- tan({{ \alpha}})tan({{ \beta}})}
\\\\\\
tan({{ \alpha}} + {{ \beta}}) = \cfrac{-\frac{8}{15}+\frac{\sqrt{11}}{5}}{1-\left( -\frac{8}{15}\cdot \frac{\sqrt{11}}{5} \right)}\implies 
\cfrac{\frac{-8+3\sqrt{11}}{15}}{1+\frac{8\sqrt{11}}{75}}
\\\\\\
tan({{ \alpha}} + {{ \beta}}) =\cfrac{\frac{-8+3\sqrt{11}}{15}}{\frac{75+8\sqrt{11}}{75}}\implies \cfrac{-8+3\sqrt{11}}{15}\cdot \cfrac{75}{75+8\sqrt{11}}
\\\\\\
$

$\bf tan({{ \alpha}} + {{ \beta}}) =\cfrac{-8+3\sqrt{11}}{1}\cdot \cfrac{3}{75+8\sqrt{11}}\implies \cfrac{15\sqrt{11}-40}{75+8\sqrt{11}}$

6 0

4 years ago