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3 years ago

What is the cosecant of angle X?

Mathematics

1 answer:

givi [52]3 years ago

8 0

Answer:

cscX = $\frac{y}{x}$

Step-by-step explanation:

using the trigonometric identity

cscX = $\frac{1}{sinX}$

sinX = $\frac{opposite}{adjacent}$ = $\frac{x}{y}$

cscX = $\frac{1}{\frac{x}{y} }$ = $\frac{y}{x}$

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H(x)= ⎩ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎧ 8+6x x+4 10 4x−3 , , , x<−7 −7≤x<−5 x≥−5

motikmotik

Answer:

$h(-6) = -\frac{1}{5}$

Step-by-step explanation:

Given

$h(x) =\left[\begin{array}{cc}8 + 6x\ \ &x < -7\\\frac{x + 4}{10} &-7 \le x$

Required

Determine h(-6)

To do this, we make use of:

$h(x) = \frac{x + 4}{10}$

Because $-7 \le -6$

So, we have:

$h(x) = \frac{x + 4}{10}$

$h(-6) = \frac{-6 + 4}{10}$

$h(-6) = \frac{-2}{10}$

Simplify

$h(-6) = -\frac{1}{5}$

4 0

3 years ago

What is the equation of a parabola if the vertex is (1, 4) and the directrix is located at y = 7?

Oksana_A [137]

According to the vertex and the directrix of the given parabola, the equation is:

$y = \frac{3}{4}(x - 1)^2 + 4$

<h3>What is the equation of a parabola given it’s vertex?</h3>

The equation of a quadratic function, of vertex (h,k), is given by:

$y = a(x - h)^2 + k$

In which a is the leading coefficient.

The directrix is at y = k + 4a.

In this problem, the vertex is (1,4), hence:

$h = 1, k = 4$

The directrix is at y = 7, hence:

$4 + 4a = 7$

$a = \frac{3}{4}$

Hence, the equation is:

$y = a(x - h)^2 + k$

$y = \frac{3}{4}(x - 1)^2 + 4$

More can be learned about the equation of a parabola at brainly.com/question/26144898

4 0

2 years ago

Enter the number in standard notation.<br> 7x 102=<br> PLEASE HELP FAST

kiruha [24]

Answer:714

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

-1/225x^2+2/3x what is the vertex of this quadratic function?

Julli [10]

$\bf \textit{vertex of a parabola}\\ \quad \\\\

\begin{array}{lccclll}
y=&-\frac{1}{225}x^2&+\frac{2}{3}x\\\\
y=&-\frac{1}{225}x^2&+\frac{2}{3}x&+0\\
&\uparrow &\uparrow &\uparrow \\
&a&b&c
\end{array}\qquad 
\left(-\cfrac{{{ b}}}{2{{ a}}}\quad ,\quad {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\right)$

6 0

3 years ago

Rewrite the expression in terms of the given function 1/1-sinx - sinx/1+sinx

Feliz [49]

Your question seems a bit incomplete, but for starters you can write

$\dfrac1{1-\sin x}-\dfrac{\sin x}{1+\sin x}=\dfrac{1+\sin x}{(1-\sin x)(1+\sin x)}-\dfrac{\sin x(1-\sin x)}{(1+\sin x)(1-\sin x)}=\dfrac{1+\sin x-\sin x(1-\sin x)}{(1-\sin x)(1+\sin x)}$

Expanding where necessary, recalling that $(1-\sin x)(1+\sin x)=1-\sin^2x=\cos^2x$ , you have

$\dfrac{1+\sin x-\sin x(1-\sin x)}{(1-\sin x)(1+\sin x)}=\dfrac{1+\sin x-\sin x+\sin^2x}{\cos^2x}=\dfrac{1+\sin^2x}{\cos^2x}$

and you can stop there, or continue to rewrite in terms of the reciprocal functions,

$\dfrac{1+\sin^2x}{\cos^2x}=\sec^2x+\tan^2x$

Now, since $1+\tan^2x=\sec^2x$ , the final form could also take

$\sec^2x+\tan^2x=\sec^2x+(\sec^2x-1)=2\sec^2x-1$

or

$\sec^2x+\tan^2x=(1+\tan^2x)+\tan^2x=1+2\tan^2x$

7 0

3 years ago