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Molodets [167]
3 years ago
9

Please help me thank you

Mathematics
1 answer:
Elena L [17]3 years ago
3 0

Answer:

option C Only second equation is an identity is correct.

Step-by-step explanation:

1)

1 +\frac{cos^2\theta}{cot^2\theta(1-sin^2\theta)}= 9 sec^2\theta

We need to prove this identity.

We know:

cos^2\theta + sin^2\theta = 1\\=> cos^2\theta = 1- sin^2\theta

and

\frac{1}{cot^2\theta } = tan^2\theta \\and\\tan^2\theta = \frac{sin^2\theta}{cos^2\theta}

Using these to solve the identity

1 +\frac{cos^2\theta}{cot^2\theta(cos^2\theta)}= 9 sec^2\theta\\1 +\frac{1}{cot^2\theta} = 9 sec^2\theta\\1+tan^2\theta = 9 sec^2\theta\\1+\frac{sin^2\theta}{cos^2\theta} = 9sec^2\theta\\\\\frac{cos^2\theta+sin^2\theta}{cos^2\theta} = 9sec^2\theta\\\frac{1}{cos^2\theta}=9sec^2\theta \\sec^2\theta \neq 9sec^2\theta

So, this is not an identity.

2)

20sin\theta(\frac{1}{sin\theta} -\frac{cot\theta}{sec\theta}) =20sin^2\theta\\

We need to prove this identity.

We know:

cot\theta = \frac{cos\theta}{sin\theta} \\and \\sec\theta=\frac{1}{cos\theta} \\so,\,\, \frac{cot\theta}{sec\theta}= \frac{\frac{cos\theta}{sin\theta}}{\frac{1}{cos\theta}}  \\Solving\\ \frac{cot\theta}{sec\theta} =\frac{cos^2\theta}{sin\theta}

Using this to solve the identity

20sin\theta(\frac{1}{sin\theta} -\frac{cot\theta}{sec\theta}) =20sin^2\theta\\\\Putting\,\,values\,\,\\ 20sin\theta(\frac{1}{sin\theta} -\frac{cos^2\theta}{sin\theta}) =20sin^2\theta\\ 20sin\theta(\frac{1-cos^2\theta}{sin\theta}) =20sin^2\theta\\1-cos^2\theta = sin^2\theta\\ 20sin\theta(\frac{sin^2\theta}{sin\theta}) =20sin^2\theta\\Cancelling\,\, sin\theta \,\,over \,\,sin\theta\\20sin^2\theta=20sin^2\theta

So, this is an identity.

So, option C Only second equation is an identity is correct.

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Answer:

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Step-by-step explanation:

We know that all the vertices of the isosceles trapezoid lie on the parabola, and the points A and D lie along the x-axis, i.e at y=0

Therefore points A and D are located where

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Now we need to find the coordinates of point C; we already have its y-coordinate (it's y=2), and looking at the figure attached we see that the x-coordinate of point C is \frac{2}{tan(60^o)} farthar from the coordinate of point C; thus

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Now this point C lies on the parabola, and therefore must satisfy the equation y=a(x+1)(x-5):

2=a(0.1547+1)(0.5147-5)

\therefore a=\frac{2}{(0.1547+1)(0.5147-5)} =-0.35747\\\\\boxed{a=-0.35747}

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