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2 years ago

11

What appears to be the scale factor used to create this dilation?

2 answers:

stealth61 [152]2 years ago

7 0

The answer is 0.5

Just take one point from the original shape and one from the transformed one. The just write the ordered pair out. Mine was (3,4) and (1.5, 2) Now in order for you to get from 3 to 1.5 you need to multiply 0.5 so the scale factor that was used is 0.5

Pls mark Braille at

MrMuchimi2 years ago

3 0

You have a lot of tabs open haha, and i think it’s .5

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Can someone help with this

xenn [34]

Answer:

1/8^2, 1/2^4 and 1/3^5

Step-by-step explanation:

3 0

3 years ago

<img src="https://tex.z-dn.net/?f=I%20%3D%20%5Cfrac%7BE%7D%7B%20%5Csqrt%7BR%20%7B%7D%5E%7B2%7D%20%2B%20W%20%7B%7D%5E%7B2%7D%20L%

AleksandrR [38]

Answer:

R = sqrt[(IWL)^2/(E^2 - I^2)] or R = -sqrt[(IWL)^2/(E^2 - I^2)]

Step-by-step explanation:

Squaring both sides of equation:

I^2 = (ER)^2/(R^2 + (WL)^2)

<=>(ER)^2 = (I^2)*(R^2 + (WL)^2)

<=>(ER)^2 - (IR)^2 = (IWL)^2

<=> R^2(E^2 - I^2) = (IWL)^2

<=> R^2 = (IWL)^2/(E^2 - I^2)

<=> R = sqrt[(IWL)^2/(E^2 - I^2)] or R = -sqrt[(IWL)^2/(E^2 - I^2)]

Hope this helps!

7 0

3 years ago

Find the volume of a cone with diamter 12 and height 8

NeX [460]

Answer:

Step-by-step explanation:

V = 1/3 * pi * r^2 * h

d = 12

r = d/2

r = 12/2

r = 6

V = 1/3 * 3.14 * 6^2 * 8

V = 301.44

3 0

3 years ago

Can someone help me find out what is 48y-24

Maksim231197 [3]

Answer: 24(2y−1)

Step-by-step explanation:

Factor 48y−24

48y−24

=24(2y−1)

Answer:

24(2y−1)

3 0

3 years ago

Read 2 more answers

Find the exact values of a) sec of theta b)tan of theta if cos of theta= -4/5 and sin<0

Gre4nikov [31]

Answer:

Using trigonometric ratio:

$\sec \theta = \frac{1}{\cos \theta}$

$\tan \theta = \frac{\sin \theta}{\cos \theta}$

From the given statement:

$\cos \theta = -\frac{4}{5}$ and sin < 0

⇒ $\theta$ lies in the 3rd quadrant.

then;

$\sec \theta = \frac{1}{-\frac{4}{5}} = -\frac{5}{4}$

Using trigonometry identities:

$\sin \theta = \pm \sqrt{1-\cos^2 \theta}$

Substitute the given values we have;

$\sin \theta = \pm\sqrt{1-(\frac{-4}{5})^2 } =\pm\sqrt{1-\frac{16}{25}} =\pm\sqrt{\frac{25-16}{25}} =\pm \sqrt{\frac{9}{25} } = \pm\frac{3}{5}$

Since, sin < 0

⇒ $\sin \theta = -\frac{3}{5}$

now, find $\tan \theta$ :

$\tan \theta = \frac{\sin \theta}{\cos \theta}$

Substitute the given values we have;

$\tan \theta = \frac{-\frac{3}{5} }{-\frac{4}{5} } = \frac{3}{5}\times \frac{5}{4} = \frac{3}{4}$

Therefore, the exact value of:

(a)

$\sec \theta =-\frac{5}{4}$

(b)

$\tan \theta= \frac{3}{4}$

7 0

3 years ago