(not sure )
my method
let projection of b onto line dq be M
BM is 2r
then let the remaining angle ( right) be K
in triangle BMK ,
MK = 29-5-r
By Pyth. theorem ,
23^2 = (2r)^2+(29-5-r)^2
529=4r^2+576-48r+r^2
5r^2-48r+47=0
r = 8.493237063 or 1.106762937(rejected)

5 0

3 years ago

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

4 years ago

Simplified form of the expression (x^3+3x^2+6)-(9x^2-5x+7)

JulijaS [17]

$(x^3 + 3x^2 + 6) - (9x^2 - 5x + 7)\\\\x^3 + 3x^2 + 6 - 9x^2 + 5x - 7\\x^3 - 6x^2 + 5x -1$

8 0

4 years ago

Think about your previous math courses. Identify one tope that you learned about which

saw5 [17]

The answer for this question I don’t think could be answered by anyone but yourself. To help you figure this out, just simply think about all of the math courses you’ve had previous to your current course and think about which course you did best in.

8 0

3 years ago