Please help, very confused!

1 answer:

7 0

$\bf cot(\theta)=\cfrac{1}{tan(\theta)}
\qquad \qquad 
csc(\theta)=\cfrac{1}{sin(\theta)}\qquad \qquad sin(-\theta )=-sin(\theta )
\\\\\\
\textit{also recall }sin^2(\theta)+cos^2(\theta)=1\implies sin^2(\theta)=1-cos^2(\theta)\\\\
-------------------------------$

$\bf \cfrac{csc^2(x)-cot^2(x)}{sin(-x)cot(x)}\implies \cfrac{\frac{1}{sin^2(x)}-\frac{cos^2(x)}{sin^2(x)}}{-sin(x)\frac{cos(x)}{sin(x)}}\implies \cfrac{\frac{1-cos^2(x)}{sin^2(x)}}{-cos(x)}
\\\\\\
\cfrac{\frac{sin^2(x)}{sin^2(x)}}{-cos(x)}\implies \cfrac{1}{-cos(x)}\implies -sec(x)$

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The area of polygon MNOPQR = Area of a rectangle that is 9 square units + Area of a rectangle that is ___ square units. (Input w

vekshin1

You have the polygon MNOPQR which can be expressed as two rectangles pasted one next to each other.

To see the two rectangles in the picture, you can draw a line parallel to segment MR througn point N.

From the original picture you can state the dimensions of both rectangles.

Call S, the point where the line that you drew intercepts the segment RQ.

Then one rectangle is MNSR and the other rectangle is OPQS.

The measures of the sides of the rectangle MNSR are:

- the length of MN = length of SR = base

- the length of MR = the length of NS.= height

So its area is base * height, which you can all A1.

The measured of the rectangle OPQS are:

- segment OP = segment SQ = segment QR - segment SR = base

- segment PQ = segment OS = height

So its area is base * height, which you can call A2.

Then the area of the polygon MNOPQRS is A1 + A2. One of them is 9 u^2 and the other is what the answer is asking for, and that you have calculated above.

With this procedure you can tell the value needed.

8 0

3 years ago

Solve the inequality |2h-3| >1

mel-nik [20]

Answer:

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