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Alja [10]
3 years ago
14

Graph XY with endpoints X(5,−2) and Y(3,−3) and its image after a reflection in the x-axis and then a rotation of 270 degrees ab

out the origin.
What are the coordinates for XY after a reflection in the x-axis and then a rotation of 270 degrees about the orgin?

Please show all the work on how you got your answer.

Mathematics
1 answer:
DerKrebs [107]3 years ago
8 0

Answer:

  X''(2, -5), Y''(3, -3)

Step-by-step explanation:

You know that reflection in the x-axis changes the sign of the y-coordinate. Points that used to be above the axis are now below by the same amount, and vice versa.

Rotation counterclockwise by 270° is the same as clockwise rotation by 90°. That maps the coordinates like this:

  (x, y) ⇒ (y, -x)

The two transformations together give you ...

  (x, y) ⇒ (x, -y) ⇒ (-y, -x) . . . . . . . . equivalent to reflection across y=-x.

Using this mapping, we have ...

  X(5, -2) ⇒ X''(2, -5)

  Y(3, -3) ⇒ Y''(3, -3) . . . . . . on the equivalent line of reflection, so invariant

_____

The attachment shows the original segment in red, the reflected segment in purple, and the rotated segment in blue. The equivalent line of reflection is shown as a dashed green line.

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Let X be the score on an english test which is normally distributed with mean of 31.5 and standard deviation of 7.3

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Here we have to find score that separates the top 59% from the bottom 41%

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This is same as finding z score such that probability below z score is 0.49 and above probability is 0.59

P(Z < z) = 0.49

Using excel function to find the z score for probability 0.49 we get

z = NORM.S.INV(0.49)

z = -0.025

It means for z score -0.025 area below it is 41% and above it is 59%

Now we will convert this z score into x value using given mean and standard deviation

x = (z* standard deviation) + mean

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Step-by-step explanation:

Given: P = $3000, r = 0.085

                A = Pe^{rt}

Where

A is the Amount

P is the Principal

r is the rate

t is the time

(a) For the amount to double, A = 2 × P

               A = 2 × $3000

               A = $6000

               6000 = 3000e^{0.085t}

               \frac{6000}{3000} = e^{0.085t}

               2 = e^{0.085t}

Take log_{e} of both sides

               log_{e}2 = log_{e}e^{0.085t}

But log_{e}e = 1

            ∴ ln2 = 0.085t

               t = \frac{ln2}{0.085}

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               A = 3 × $3000

               A = $9000

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               \frac{9000}{3000} = e^{0.085t}

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               log_{e}3 = log_{e}e^{0.085t}

But log_{e}e = 1

            ∴ ln3 = 0.085t

               t = \frac{ln3}{0.085}

               t = \frac{1.0986}{0.085}

               t = 12.92

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