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3 years ago

Plz help you have to find the angle marked x in each of these polygons

Mathematics

1 answer:

svetoff [14.1K]3 years ago

5 0

Https://us-static.z-dn.net/files/dd0/5d5ddccb833d70732c67b84ec119f42f.jpeg

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Verify the identity (tan x + 1)^2 + (tan x-1)^2= 2 sec^2 x

Elina [12.6K]

$(\tan x+1)^2+(\tan x-1)^2=2\sec^2x\\\\\text{use}\ \tan x=\dfrac{\sin x}{\cos x}\\\\L_s=\left(\dfrac{\sin x}{\cos x}+\dfrac{\cos x}{\cos x}\right)^2+\left(\dfrac{\sin x}{\cos x}-\dfrac{\cos x}{\cos x}\right)^2\\\\=\left(\dfrac{\sin x+\cos x}{\cos x}\right)^2+\left(\dfrac{\sin x-\cos x}{\cos x}\right)^2\\\\=\dfrac{(\sin x+\cos x)^2}{\cos^2x}+\dfrac{(\sin x-\cos x)^2}{\cos^2x}\\\\\text{use}\ (a\pm b)^2=a^2\pm2ab+b^2$

$=\dfrac{\sin^2x+2\sin x\cos x+\cos^2}{\cos^2x}+\dfrac{\sin^2x-2\sin x\cos x+\cos^2}{\cos^2x}\\\\=\dfrac{\sin^2x+2\sin x\cos x+\cos^2+\sin^2x-2\sin x\cos x+\cos^2}{\cos^2x}\\\\=\dfrac{2\sin^2x+2\cos^2x}{\cos^2x}=\dfrac{2(\sin^2x+\cos^2x)}{\cos^2x}\\\\\text{use}\ \sin^2x+\cos^2x=1\\\\=\dfrac{2(1)}{\cos^2x}=2\cdot\dfrac{1}{\cos^2x}=2\left(\dfrac{1}{\cos x}\right)^2\\\\\text{use}\ \sec x=\dfrac{1}{\cos x}\\\\=2(\sec^2x)=2\sec^2x=R_s\\\\L_s=R_s\Rightarrow The\ identity$

4 0

3 years ago

Simplify (-4x^2-3y-8)+(5y-2x^2+1)

ddd [48]

Add the like terms.

The like terms in this case are:

-4x^2 and -2x^2

This makes -6x^2

-3y and 5y

This makes 2y

-8 and 1

This makes -7

Your answer is (-6x^2+2y-7)

8 0

2 years ago

Put in order from least to greatest: quart, pint, gallon, liter, cup

cricket20 [7]

From least to greatest
1) cup
2) pint
3) quart
4) liter
5) gallon

4 0

3 years ago

Find the gcf of <br> 24, 14, 21

kobusy [5.1K]

Answer:

Step-by-step explanation:

All the numbers provided can have a factor of 1.

eg.)

1*24=24

1*14=14

1*21=21

This is because there is only one factor they all commonly share, 1.

4 0

3 years ago

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DochEvi [55]

Answer:

The answer is C 40/3

Hope I Helped

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3 years ago

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