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3 years ago

Show that each statement is true: Cos^4X-sin^4X=2cos^2X-1

1 answer:

Ivan3 years ago

4 0

$cox^4x-sin^4x=2cos^2x-1\\\\
(cosx^2-sin^2x)(cosx^2x+sin^2x)=2cos^2x-1\\\\
cosx^2-sin^2x*1=2cos^2x-1\ \ \ | subtract\ 2cos^2x\\\\
-cosx^2-sin^2x=-1\ \ \ | multiply\ by\ -1\\\\
cosx^2+sin^2x=1\\\\
1=1$

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A given shape that is bounded by three sides and has got three internal angles is referred to as a triangle. Thus the value of PB is 8.0 units.

A given shape that is bounded by three sides and has got three internal angles is referred to as a triangle. Types of triangles include right angle triangle, isosceles triangle, equilateral triangle, acute angle triangle, etc. The sum of the internal angles of any triangle is $180^{o}$ .

In the given question, point P is center of the given triangle such that <APB = <APC = <BPC = $120^{o}$ . Such that line PB bisects <ABC into two equal measures of $30^{o}$ .

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Apply the Sine rule to determine the value of PB, such that;

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2 years ago

Let X be the temperature in at which a certain chemical reaction takes place, and let Y be the temperature in (so Y = 1.8X + 32)

Black_prince [1.1K]

Answer:

See explanation

Step-by-step explanation:

Solution:-

The random variable, Y be the temperature of chemical reaction in degree fahrenheit be a linear expression of a random variable X : The temperature in at which a certain chemical reaction takes place.

Y = 1.8*X + 32

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P ( X ≤ η ) = 0.5

- Then the median of "Y" distribution can be expressed with the help of the relation given:

P ( Y ≤ 1.8*η + 32 )

- The left hand side of the inequality can be replaced by the linear relation:

P ( 1.8*X + 32 ≤ 1.8*η + 32 )

P ( 1.8*X ≤ 1.8*η ) ..... Cancel "1.8" on both sides.

P ( X ≤ η ) = 0.5 ...... Proven

Hence,

- Through conjecture we proved that: (1.8*η + 32) has to be the median of distribution "Y".

- Recall that the definition of proportion (p) of distribution that lie within the 90th percentile. It can be mathematically expressed as the probability of random variate "X" at 90th percentile :

P ( X ≤ p_.9 ) = 0.9 ..... 90th percentile

- Now use the conjecture given as a linear expression random variate "Y",

P ( Y ≤ 1.8*p_0.9 + 32 ) = P ( 1.8*X + 32 ≤ 1.8*p_0.9 + 32 )

= P ( 1.8*X ≤ 1.8*p_0.9 )

= P ( X ≤ p_0.9 )

= 0.9

- So from conjecture we saw that the 90th percentile of "X" distribution is also the 90th percentile of "Y" distribution.

- The more general relation between two random variate "Y" and "X" is given:

Y = aX + b

Where, a : is either a positive or negative constant.

- Denote, (np) as the 100th percentile of the X distribution, so the corresponding 100th percentile of the Y distribution would be : (a*np + b).

- When a is positive,

P ( Y ≤ a*p_% + b ) = P ( a*X + b ≤ a*p_% + b )

= P ( a*X ≤ a*p_% )

= P ( X ≤ p_% )

= np_%

- When a is negative,

P ( Y ≤ a*p_% + b ) = P ( a*X + b ≤ a*p_% + b )

= P ( a*X ≤ a*p_% )

= P ( X ≥ p_% )

= 1 - np_%

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