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zvonat [6]
2 years ago
7

What’s the correct answer for this ? Two chords AB and CD intersect at E. If AE = 2cm, EB =4, and CE = 2.5 cm, find the length o

f ED
Mathematics
1 answer:
ss7ja [257]2 years ago
7 0

Answer:

ED = 3.2 cm

Step-by-step explanation:

According to chord-chord power theorem,

(AE)(EB) = (CE)(ED)

2*4 = 2.5 *ED

8/2.5 = ED

ED = 3.2 cm

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212 divided by 8 is 26.5 which would be C. I hope this helps!
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3 years ago
Mr. Paulsen challenges Olivia and Angelica to move figure ABCDE onto figure A"B"C"D"E" using a series of two different transform
Jobisdone [24]

Problem 1

One method Olivia could have used is to reflect over the x axis and then translate 10 units to the right.

Let's apply these set of steps to point C.

C = (-2,-4)

C' = (-2,4) after reflecting over the x axis; y coordinate flips in sign

C'' = (8,4) after shifting to the right 10 units; add 10 to the x coordinate

Point C(-2,-4) moves to C''(8,4) which is shown in the diagram. I'll let you check the other four points.

-----------

Another method Olivia could have used is to shift to the right 10 units first, then reflect over the x axis. In this case, the order doesn't matter. Though for some other combinations of transformations, order does matter.

============================================================

Problem 2

Reflecting over the y axis will have the point (x,y) turn into (-x,y). Only the x coordinate changes, and we flip the sign here. The y coordinate stays the same.

A point like C(-2,-4) becomes C'(2,-4) after the y axis reflection. Shifting up 8 units means we add 8 to the y coordinate to arrive at C''(2,4) but this does not match what the diagram shows. Therefore, Angelica's method is incorrect.

You should find other inconsistencies for the other four points as well.

============================================================

Problem 3

Michael's method isn't correct either. He does not apply an x axis reflection and instead relies entirely on translations to move ABCDE to A''B''C''D''E''

At a quick glance, it looks like Michael may be correct. But upon a much closer look, you'll see that things don't line up perfectly.

Instead of point C, we'll pick on point D this time.

D is located at (-6,-2). If we shift up 8 units, then we add 8 to the y coordinate to get to D'(-6,6). Then add 10 to the x coordinate to shift 10 units to the right. This leads us to D''(4,6).

However, the diagram shows D'' is actually at (4,2)

I'll let you check the other points. You should find that only points C and E go to the correct locations, but the other points do not. Again, this is because Michael did not apply an x axis reflection. His figure of A''B''C''D''E'' is upside down compared to what it should be.

============================================================

To summarize, the methods described in problem 1 are the correct way to get ABCDE to move to A''B''C''D''E''. If Olivia followed either of those two methods, then she is correct. Both Angelica and Michael have incorrect methods.

5 0
2 years ago
HELP ASAP!!!!! Given: Point B is on the perpendicular bisector of AC¯¯¯¯¯. BD¯¯¯¯¯ bisects AC¯¯¯¯¯ at point D. Prove: B is equid
Harlamova29_29 [7]

Answer:  Missing parts are,

In first blank,  AD\cong DC,

In second blank, SAS postulate

In third blank, CPCTC postulate

Step-by-step explanation:

Since, Here D is the mid point on the line segment AC.

And BD is a perpendicular to the line AC.

Therefore, In triangles ADB and CDB ( shown in figure)

AD\cong DC ( By the definition of mid point)

\angle BDA\cong \angle BDC ( right angles )

BD\cong BD ( reflexive)

Thus, By SAS ( side angle side )postulate,

\triangle ADB\cong \triangle CDB

So, by CPCTC( Corresponding parts of congruent triangles are congruent)

AB\cong CB

Now, By definition of congruent segment,

AB=CB

By definition of equidistant,

B is equally far from both A and C.




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tatuchka [14]
<h3><em><u>Answer:-6 divided by ( - 2/3) times(-5)</u></em></h3>

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