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Mekhanik [1.2K]
2 years ago
14

Please fix the question and help me do it the right way

Mathematics
1 answer:
Anestetic [448]2 years ago
3 0

the last step is wrong because we multiply the powers in this case we do not subtract them.

we use power subtract in division.

we use power multiplication when we raise to power and the root is power like square root of x

\sqrt{x}  =  {x}^{ \frac{1}{2} }

so

\sqrt[3]{x}  =  {x}^{ \frac{1}{3} }

in our case

this step is wrong

2x \times  \sqrt[3]{ \frac{2}{y}  \times  \frac{y}{y} }

it should be

2 x \times  \sqrt[3]{ \frac{2}{y}  \times  \frac{ {y}^{2} }{ {y}^{2} } }

so we get

2x \times  \sqrt[3]{2 {y}^{2} \times  \frac{1}{ {y}^{3} }  }

=  \frac{2x}{y}  \times  \sqrt[3]{2 {y}^{2} }

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The angle θ lies in Quadrant II. sinθ=3/4 What is cosθ ?
Elden [556K]

Hello from MrBillDoesMath

Answer:

[email protected] = - sqrt(7)/ 4    

which is choice B

Discussion:

This problem can be solved by drawing triangles and looking at ratios of sides or by using the trig identity:

([email protected])^2 + (sin2)^2 = 1

If [email protected] = 3/4 , the

([email protected])^2 + (3/4)^2 = 1 =>              (subtract (3/4)^2 from both sides)

([email protected])^2 = 1 - (3/4)^2  = 1 - 9/16   = 7/16

So...... taking the square root of both sides gives

[email protected] =  +\- sqrt(7)/ sqrt(16) = +\- sqrt(7)/4

But is [email protected] positive or negative?  We are told that @ is in the second quadrant and cos(@) is negative in this quadrant, so our answer must be negative

[email protected] = - sqrt(7)/ 4    

which is choice B


Thank you,

Mr. B


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Let c be the amount of parts the student answered correctly.  Suppose that s(c) is a function between c and the score of a student's project.  As the student initially receives a fixed 50 points for turning the project in and 5 points for each correct part, the function is initially:

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Answer:

0

Step-by-step explanation:

Hello, dividing by 0 is not defined. so

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This being said, we can simplify by 2x

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Thank you

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