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dusya [7]
3 years ago
8

which of the following is a solution of 2cos^2x-cost-1=0 a. 0 degrees b. 150 degrees c. 180 degrees d. 210 degrees

Mathematics
1 answer:
Luba_88 [7]3 years ago
3 0

Answer:

{0, 150} degrees

Step-by-step explanation:

Given 2cos^2x-cost-1=0, let's simplify this problem by temporarily replacing cos x with y:

2y^2 - y -1 = 0

This can be solved by factoring:  (2y + 1)(y - 1) = 0.  From this we get two solutions:  y = -1/2 and y = 1.

Remembering that we let y = cos x, we now solve:

cos x = -1/2 and cos x = 1.

Note that cos x = 1 when x = 0 and the "adjacent side" coincides with the hypotenuse.

cos x = -1/2 when the hypotenuse is 2 and the "adjacent side" is -1.  This has two solutions between 0 and 360 degrees:  150 degrees and  270 degrees.

Four answer choices are given.  Both (a) (0 degrees) and (b) (150 degrees) satisfy the original equation.  Thus, the solution set is {0, 150} (degrees).

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Step-by-step explanation:

From the question we are told that

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Generally the probability that a boy born full term in the US weighs less than 7 lbs is mathematically represented as

     P( X  <  7 ) =  P( \frac{X - \mu }{\sigma}  <  \frac{7 - 7.7 }{1.1 } )

\frac{X -\mu}{\sigma }  =  Z (The  \ standardized \  value\  of  \ X )

=>   P( X  <  7 ) =  P(Z

From the z table  the area under the normal curve to the left corresponding to    -0.6364  is

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       \sigma_{x} = \frac{\sigma }{\sqrt{n} }

=>     \sigma_{x} = \frac{ 1.1  }{\sqrt{25} }

=>     \sigma_{x} = 0.22

Generally the probability that the average weight of 25 baby boys born in a particular hospital have an average weight less than 7 lbs is mathematically represented as

         P( \= X  <  7 ) =  P( \frac{\= X - \mu }{\sigma_{x}}  <  \frac{7 - 7.7 }{ 0.22 } )

=>      P( X  <  7 ) =  P(Z

From the z table  the area under the normal curve to the left corresponding to    -3.1818  is

=>      P(Z

=>      P( X  <  7 ) = 0.00073

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