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3 years ago

No Tessa and Kari are swimming

Mathematics

2 answers:

hichkok12 [17]3 years ago

8 0

Um....Yes!

~ItsOniiSama ❤

Hope this helps...

Plz Mark brainliest

Mamont248 [21]3 years ago

7 0

Answer:

what is the question?

Step-by-step explanation:

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Use a half-angle identity to find the exact value

Tatiana [17]

Given:

$\cos 15^{\circ}$

To find:

The exact value of cos 15°.

Solution:

$$\cos 15^{\circ}=\cos\frac{ 30^{\circ}}{2}$

Using half-angle identity:

$$\cos \left(\frac{x}{2}\right)=\sqrt{\frac{1+\cos (x)}{2}}$

$$\cos \frac{30^{\circ}}{2}=\sqrt{\frac{1+\cos \left(30^{\circ}\right)}{2}}$

Using the trigonometric identity: $\cos \left(30^{\circ}\right)=\frac{\sqrt{3}}{2}$

$$=\sqrt{\frac{1+\frac{\sqrt{3}}{2}}{2}}$

Let us first solve the fraction in the numerator.

$$=\sqrt{\frac{\frac{2+\sqrt{3}}{2}}{2}}$

Using fraction rule: $\frac{\frac{a}{b} }{c}=\frac{a}{b \cdot c}$

$$=\sqrt{\frac {2+\sqrt{3}}{4}}$

Apply radical rule: $\sqrt[n]{\frac{a}{b}}=\frac{\sqrt[n]{a}}{\sqrt[n]{b}}$

$$=\frac{\sqrt{2+\sqrt{3}}}{\sqrt{4}}$

Using $\sqrt{4} =2$ :

$$=\frac{\sqrt{2+\sqrt{3}}}{2}$

$$\cos 15^\circ=\frac{\sqrt{2+\sqrt{3}}}{2}$

5 0

3 years ago

What’s the slope of 10,6 and 4, 1.2

Kruka [31]

let's firstly change the 1.2 to a fraction

$1.\underline{2}\implies \cfrac{12}{1\underline{0}}\implies \cfrac{6}{5} \\\\[-0.35em] ~\dotfill\\\\ (\stackrel{x_1}{10}~,~\stackrel{y_1}{6})\qquad (\stackrel{x_2}{4}~,~\stackrel{y_2}{\frac{6}{5}}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{\frac{6}{5}}-\stackrel{y1}{6}}}{\underset{run} {\underset{x_2}{4}-\underset{x_1}{10}}}\implies \cfrac{~~ \frac{6-30}{5}~~}{-6}\implies \cfrac{~~ \frac{-24}{5}~~}{-6}\implies \cfrac{~~ -\frac{24}{5}~~}{-\frac{6}{1}}$

$-\cfrac{\stackrel{4}{~~\begin{matrix} 24 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}}{5}\cdot -\cfrac{1}{\underset{1}{~~\begin{matrix} 6 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}}\implies \boxed{\cfrac{4}{5}}$