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oksano4ka [1.4K]

3 years ago

9

Find the image of A(6, -4) after it is reflected over the line y = - 2, then reflected over the line x = 1. (-4, 6) (-4, 0) (-4,

-4) (6, 2)

1 answer:

olchik [2.2K]3 years ago

7 0

Answer:(-4,0)

Step-by-step explanation:

Given

Point A (6,-4)

Reflection of Point A about line y=-2 is

A' (6,0)

If any Point is reflected about any line y=constant then its x coordinate remains same

Now reflection of Point A' about x=1 is

A'' (-4,0)

If any point is reflected about line x=constant then its y coordinate remains same

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A lighthouse is located on an island 33 miles from the closest point on a straight shoreline. If the lighthouse light rotates cl

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Question:

A lighthouse is located on an island 3 miles from the closest point on a straight shoreline. If the lighthouse light rotates clockwise at a constant rate of 9 revolutions per minute, how fast does the beam of light move towards the point on the shore closest to the island when it is 52 miles from that point

Answer:

The beam of light moves at $16278\pi$ miles/min

Step-by-step explanation:

This question is illustrated with the attached image

Taking the instructions in the question, one at a time.

A revolution of 9 per minute implies that:

$\frac{d\theta}{dt} = \frac{9 * 2\pi\ rad}{1\ min}$

$\frac{d\theta}{dt} = 18\pi \frac{rad}{min}$

Take tan of the angle in the attachment:

$tan(\theta) =\frac{52}{3}$

Differentiate both sides with respect to time

$\frac{d\ tan(\theta)}{dt} =\frac{52}{3} * \frac{dx}{dt}$

Rewrite as:

$\frac{d\ tan(\theta)}{d\theta} * \frac{d\theta}{dt} =\frac{52}{3} * \frac{dx}{dt}$

In calculus:

$sec^2(\theta) =\frac{d\ tan(\theta)}{d\theta}$ -- Chain rule

So:

$sec^2(\theta) *\frac{d\theta}{dt} =\frac{52}{3} * \frac{dx}{dt}$

In trigonometry:

$sec^2(\theta) = tan^2(\theta) + 1$

So:

$(tan^2(\theta) + 1)\frac{d\theta}{dt} =\frac{52}{3} * \frac{dx}{dt}$

Recall that:

$tan(\theta) =\frac{x}{3}$

$((\frac{52}{3})^2 + 1)\frac{d\theta}{dt} =\frac{1}{3} * \frac{dx}{dt}$

$(\frac{52^2}{9} + 1)\frac{d\theta}{dt} =\frac{1}{3} * \frac{dx}{dt}$

$(\frac{2704}{9} + 1)\frac{d\theta}{dt} =\frac{1}{3} * \frac{dx}{dt}$

$(\frac{2704+9}{9})\frac{d\theta}{dt} =\frac{1}{3} * \frac{dx}{dt}$

$(\frac{2713}{9})\frac{d\theta}{dt} =\frac{1}{3} * \frac{dx}{dt}$

Recall that: $\frac{d\theta}{dt} = 18\pi \frac{rad}{min}$

$(\frac{2713}{9}) * 18\pi =\frac{1}{3} * \frac{dx}{dt}$

$2713 * 2\pi =\frac{1}{3} * \frac{dx}{dt}$

Multiply both sides by 3

$3 * 2713 * 2\pi =\frac{1}{3} * \frac{dx}{dt} * 3$

$3 * 2713 * 2\pi =\frac{dx}{dt}$

$16278\pi =\frac{dx}{dt}$

$\frac{dx}{dt} = 16278\pi$ miles/min

Hence:

The beam of light moves at $16278\pi$ miles/min

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Solve the equation 0= x^2 + 4x– 12 by graphing.

DiKsa [7]

Answer:

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Step-by-step explanation:

${x}^{2} + 4x - 12 = 0 \\ 4 \div 2 = 2 \\ x1 = - 2 - u \\ x2 = - 2 + u \\ x1x2 = ( - 2 - u)( - 2 + u) = - 12 \\ 4 - {u}^{2} = - 12 \\ 4 + 12 = {u}^{2} \\ 16 = {u}^{2} \\ \sqrt{16} = + \: or - \sqrt{ {u}^{2} } \\ u = + or - 4 \\$

$x1 = - 2 - 4 = - 6 \\ x2 = - 2 + 4 = 2$

8 0

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