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2 years ago

Pleeease help meeeeee

Mathematics

1 answer:

bixtya [17]2 years ago

8 0

Answer:

Step-by-step explanation:

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Let event A= the place is a city let event B= the place in north America which outcomes are in A or B?

fredd [130]

Answer:

Step-by-step explanation:

Given two events which are

A = the event that the place is a city

and

B = the event that the place is in north America.

We have to find outcomes which are either A or B

This can be denoted as event AUB which implies A or B or both

The event A or B is defined by the place is either a city or it is in north America or a place in north America.

Hence AUB = the place is a city or place is in North America

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2 years ago

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Find the area of this shape<br> giving brainliest please help

pogonyaev

ANSWER: ⇒ 366.19cm²

I hope this helps you :)

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2 years ago

What the freak is this 4^(x)-2^(x+1)=48

Dmitry_Shevchenko [17]

Answer:

4^x - 2^(x+1) = 48

x=3

Step-by-step explanation:

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3 years ago

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On a graph of the area of a square vs. its perimeter, a few points are plotted.

k0ka [10]

Answer:

Step-by-step explanation:

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2 years ago

Prove that sin3a-cos3a/sina+cosa=2sin2a-1

Sloan [31]

Answer:

$\frac{sin(3a)-cos(3a)}{sin(a)+cos(a)} =2sin(2a)-1$

Step-by-step explanation:

we are given

$\frac{sin(3a)-cos(3a)}{sin(a)+cos(a)} =2sin(2a)-1$

we can simplify left side and make it equal to right side

we can use trig identity

$sin(3a)=3sin(a)-4sin^3(a)$

$cos(3a)=4cos^3(a)-3cos(a)$

now, we can plug values

$\frac{(3sin(a)-4sin^3(a))-(4cos^3(a)-3cos(a))}{sin(a)+cos(a)}$

now, we can simplify

$\frac{3sin(a)-4sin^3(a)-4cos^3(a)+3cos(a)}{sin(a)+cos(a)}$

$\frac{3sin(a)+3cos(a)-4sin^3(a)-4cos^3(a)}{sin(a)+cos(a)}$

$\frac{3(sin(a)+cos(a))-4(sin^3(a)+cos^3(a))}{sin(a)+cos(a)}$

now, we can factor it

$\frac{3(sin(a)+cos(a))-4(sin(a)+cos(a))(sin^2(a)+cos^2(a)-sin(a)cos(a)}{sin(a)+cos(a)}$

$\frac{(sin(a)+cos(a))[3-4(sin^2(a)+cos^2(a)-sin(a)cos(a)]}{sin(a)+cos(a)}$

we can use trig identity

$sin^2(a)+cos^2(a)=1$

$\frac{(sin(a)+cos(a))[3-4(1-sin(a)cos(a)]}{sin(a)+cos(a)}$

we can cancel terms

$=3-4(1-sin(a)cos(a))$

now, we can simplify it further

$=3-4+4sin(a)cos(a))$

$=-1+4sin(a)cos(a))$

$=4sin(a)cos(a)-1$

$=2\times 2sin(a)cos(a)-1$

now, we can use trig identity

$2sin(a)cos(a)=sin(2a)$

we can replace it

$=2sin(2a)-1$

so,

$\frac{sin(3a)-cos(3a)}{sin(a)+cos(a)} =2sin(2a)-1$

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2 years ago

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