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3 years ago

Solve 2csc^2x-2cscx-1=0 for 0° ≤ x ≤ 360°

Mathematics

1 answer:

umka21 [38]3 years ago

6 0

Answer:

Step-by-step explanation:

2 csc²x-2 csc x-1=0

$\frac{2}{sin^2x} -\frac{2}{sin x} -1=0\\multiply~by~sin^2x\\2-2sin~x-sin^2x=0\\sin^2x+2sinx-2=0\\sin ~x=\frac{-2 \pm\sqrt{2^2-4*1*(-2)} }{2*1} \\=\frac{-2 \pm\sqrt{4+8} }{2} \\=\frac{-2 \pm\sqrt{4*3} }{2} \\=\frac{-2 \pm2\sqrt{3} }{2} \\=-1\pm\sqrt{3} \\|sin~x| \le ~1\\so~sin~x=-1+\sqrt{3} \\sin~x=\sqrt{3} -1\\x=sin^{-1}(\sqrt{3} -1) \approx 47.06^\circ,180-47.06\\or~x=47.06^\circ,132.94^\circ$

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Identify an equation in point-slope form for the line perpendicular to y = -1/2x + 11 that passes through (4, -8)

andrey2020 [161]

Answer:

y+8 = 2(x-4)

Step-by-step explanation:

Perpendicular lines have negative reciprocal slopes. Since we know the slope of the first line is -1/2, this means the slope of the line perpendicular to it is 2.

And since we know the coordinated located on the line, we can fill out the formula for poiny-slope form:

y-y1 = m(x-x1)

6 0

3 years ago

In a circle a chord 10 inches long is parallel to a tangent and bisects the radius drawn to the point of tangency. How long is t

Lorico [155]

Check the picture below.

keeping in mind that the point of tangency for a radius line and a tangent is alway a right-angle, since the "red" chord is parallel to the "green" tangent line outside, then the chord is cutting the "green" radius there in two equal halves at a right-angle, as you see in the picture.

we know the chord is 10 units long, so 5 + 5, since is perpendicularity with the radius will also cut the chord in two equal halves.

anyhow, all that said, we end up with triangle you see on the right-hand-side, and then we can just use the pythagorean theorem.

$\bf \textit{using the pythagorean theorem}
\\\\
c^2=a^2+b^2
\qquad 
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
-------\\
c=2a\\
a=a\\
b=5
\end{cases}\implies (2a)^2=a^2+5^2$

$\bf (2^2a^2)=a^2+25\implies 4a^2=a^2+25\implies 3a^2=25
\\\\\\
a^2=\cfrac{25}{3}\implies a=\sqrt{\cfrac{25}{3}}\implies a=\cfrac{\sqrt{25}}{\sqrt{3}}\implies a=\cfrac{5}{\sqrt{3}}
\\\\\\
\textit{and then we can \underline{rationalize the denominator} }
\\\\\\
a=\cfrac{5}{\sqrt{3}}\cdot \cfrac{\sqrt{3}}{\sqrt{3}}\implies a=\cfrac{5\sqrt{3}}{\sqrt{3^2}}\implies a=\cfrac{5\sqrt{3}}{3}$

7 0

4 years ago

Plz help asap ssjfjg

Finger [1]

Answer:

Step-by-step explanation:

3+6+x+4+3=94

16+x=94

x=78

The mode is 4

7 0

4 years ago

Find the slope of the line. A: 1/2 B: - 1/2 C: -2 D: 2 Please Help

notka56 [123]

It is B) -1/2 you btw you don't have to put please help :)

4 0

4 years ago

GET BRAINLIEST FOR THE CORRECT ANSWER AND EXPLANATION!

Tresset [83]

Interpreting the inequality, it is found that the correct option is given by F.

------------------

The first equation is of the line.
The equal sign is present in the inequality, which means that the line is not dashed, which removes option G.

In standard form, the equation of the line is:

$x + 2y = 6$

$2y = 6 - x$

$y = -0.5x + 2$

Thus it is a decreasing line, which removes options J.

We are interested in the region on the plane below the line, that is, less than the line, which removes option H.

------------------

As for the second equation, the normalized equation is:

$3x^2 + 3y^2 = 12$

$3(x^2 + y^2) = 12$

$x^2 + y^2 = 4$

Thus, a circle centered at the origin and with radius 2.
Now, we have to check if the line $x + 2y - 6 = 0$ , with coefficients $a = 1, b = 2, c = -6$ , intersects the circle, of centre $x = 0, y = 0$
First, we find the following distance:

$d = \sqrt{\frac{|ax + by + c|}{a^2 + b^2}}$

Considering the coefficients of the line and the center of the circle.

$d = \sqrt{\frac{|1(0) + 2(0) - 6|}{1^2 + 2^2}} = \sqrt{\frac{6}{5}} = 1.1$

This distance is less than the radius, thus, the line intersects the circle, which removes option K, and states that the correct option is given by F.

A similar problem is given at brainly.com/question/16505684

3 0

3 years ago