$\sin(60°) = \frac{x}{14} \Leftrightarrow x = 14 \sin(60°) = \frac{14 \sqrt{3} }{2} = 7 \sqrt{3} \\ \cos(60°) = \frac{y}{14} \Leftrightarrow y = 14\cos(60°) = \frac{14}{2} = 7$

6 0

3 years ago

Help plz I don't understand

Vedmedyk [2.9K]

You would multiply 36 and 6 to get x

5 0

4 years ago

If a and b are the roots of the quadratic equation 2r + 6x -7 = 0, form the equation with the following causes.

Pepsi [2]

Answer:

Step-by-step explanation:

Assuming the given equation is actually $2x^2+6x-7=0$ .

Then $\alpha +\beta=-\frac{6}{2} =-3,and,\alpha \beta=-\frac{7}{2}$

The sum of roots of the new equation is $\frac{1}{2\alpha+1} +\frac{1}{2\beta +1} =\frac{2(\alpha+\beta+1)}{4\alpha \beta+2(\alpha +\beta)+1}$

$\implies \frac{1}{2\alpha+1} +\frac{1}{2\beta +1} =\frac{2(-3+1)}{4*-3.5+2(-3)+1} =\frac{4}{19}$

The product of the roots of the new equation is $\frac{1}{2\alpha +1}*\frac{1}{2\beta+1}=\frac{1}{4\alpha \beta+2(\alpha+\beta)+1}$

$\implies \frac{1}{2\alpha +1}*\frac{1}{2\beta+1}=\frac{1}{4*-3.5+2(-3)+1} =-\frac{1}{19}$

The new equation is given by:

$x^2-(sum\:of\:roots)x+product\:of\:roots=0$

$x^2-\frac{4}{19} x-\frac{1}{19} =0$

$19x^2-x-1=0$

6 0

4 years ago