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3 years ago

6

A dilation has a center (0,0,0), Find the image fo the point (-1, -2, 0) for the scale factor of 3.

1 answer:

pychu [463]3 years ago

7 0

Answer:

(- 3, - 6, 0)

Step-by-step explanation:

Since the dilatation is centred at the origin, multiply each of the coordinates by the scale factor 3

(- 1, - 2, 0) → (- 3, - 6, 0)

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Can someone explain how to solve this and the solution? Please and thank you :)

Verizon [17]

25 > 2x + 8 - 66

first simplify all numbers

25 > 2x + (8 - 66)
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25 (+58) > 2x - 58 (+58)
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3 years ago

Quadrilateral ABCD is located at A(-2,2), B(-2,4) and D(2,2). The quadrilateral is then transformed using the rule (x+2,y-3) to

Kay [80]

The rule is x+2, y-3 so you apply it to each of these coordinates. Right 2, down 3.
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7 0

3 years ago

You stand a known distance from the base of the tree, measure the angle of elevation the top of the tree to be 15â—¦ , and then

gogolik [260]

Answer:

The maximum possible error of in measurement of the angle is $d\theta_1 =(14.36p)^o$

Step-by-step explanation:

From the question we are told that

The angle of elevation is $\theta_1 = 15 ^o = \frac{\pi}{12}$

The height of the tree is h

The distance from the base is D

h is mathematically represented as

$h = D tan \theta$ Note : this evaluated using SOHCAHTOA i,e

$tan\theta = \frac{h}{D}$

Generally for small angles the series approximation of $tan \theta \ is$

$tan \theta = \theta + \frac{\theta ^3 }{3}$

So given that $\theta = 15 \ which \ is \ small$

$h = D (\theta + \frac{\theta^3}{3} )$

$dh = D (1 + \theta^2) d\theta$

=> $\frac{dh}{h} = \frac{1 + \theta ^2}{\theta + \frac{\theta^3}{3} } d \theta$

Now from the question the relative error of height should be at most

$\pm p$ %

=> $\frac{dh}{h} = \pm p$

=> $\frac{1 + \theta ^2}{\theta + \frac{\theta^3}{3} } d \theta = \pm p$

=> $d\theta = \pm \frac{\theta + \frac{\theta^3}{3} }{1+ \theta ^2} * \ p$

So for $\theta_1$

$d\theta_1 = \pm \frac{\theta_1 + \frac{\theta^3_1 }{3} }{1+ \theta_1 ^2} * \ p$

substituting values

$d [\frac{\pi}{12} ] = \pm \frac{[\frac{\pi}{12} ] + \frac{[\frac{\pi}{12} ]^3 }{3} }{1+ [\frac{\pi}{12} ] ^2} * \ p$

=> $d\theta_1 = 0.25 p$

Converting to degree

$d\theta_1 = (0.25* 57.29) p$

$d\theta_1 =(14.36p)^o$

4 0

3 years ago