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photoshop1234 [79]
3 years ago
10

Please help !! thanks in advance :D

Mathematics
2 answers:
omeli [17]3 years ago
6 0
CIRCLES

◆ A little manipulation of the diagram
( as shown in the attachment ) 

Once you're done with this simple manipulation and the labeling, we get the equation : OB² + BC² = OC²

 • r² = ( r - 9 )² + 15² 

Solve this trivial equation to get : r = 17 

 • Radius = 17 is your final answer .
_____________________________________________________________

Hope this helps you :)

Whitepunk [10]3 years ago
5 0

Answer:

<h2>a) 17</h2>

Step-by-step explanation:

<em>Look at the picture.</em>

We have the right triangle (red). Use the Pythagorean theorem:

leg^2+leg^2=hypotenuse^2

We have

leg=15,\ leg=r-9,\ hypotenuse=r

Substitute:

15^2+(r-9)^2=r^2       <em>use (a - b)² = a² - 2ab + b²</em>

225+r^2-2(r)(9)+9^2=r^2

225+r^2-18r+81=r^2         <em>subtract r² from both sides</em>

306-18r=0          <em>subtract 306 from both sides</em>

-18r=-306      <em>divide both sides by (-18)</em>

r=17

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vichka [17]

Answer:

(0, -1)

Step-by-step explanation:

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Since we still have a fraction, we shall do it one more time. This time we multiply by 3

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2 years ago
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timurjin [86]

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The coordinates of the vertices of ΔPQR are (–2, –2), (–6, –2), and (–6, –5).
melamori03 [73]
<h3>Answer:</h3>

Yes, ΔPʹQʹRʹ is a reflection of ΔPQR over the x-axis

<h3>Explanation:</h3>

The problem statement tells you the transformation is ...

... (x, y) → (x, -y)

Consider the two points (0, 1) and (0, -1). These points are chosen for your consideration because their y-coordinates have opposite signs—just like the points of the transformation above. They are equidistant from the x-axis, one above, and one below. Each is a <em>reflection</em> of the other across the x-axis.

Along with translation and rotation, <em>reflection</em> is a transformation that <em>does not change any distance or angle measures</em>. (That is why these transformations are all called "rigid" transformations: the size and shape of the transformed object do not change.)

An object that has the same length and angle measures before and after transformation <em>is congruent</em> to its transformed self.

So, ... ∆P'Q'R' is a reflection of ∆PQR over the x-axis, and is congruent to ∆PQR.

6 0
3 years ago
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