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DiKsa [7]
2 years ago
5

Prove that the medians to the legs of an isosceles triangle are congruent.

Mathematics
1 answer:
Vilka [71]2 years ago
8 0

Step-by-step explanation:

Let ABC be an isosceles triangle with sides AC and BC of equal length.                

We need to prove that the medians AD and BE are of equal length.

Consider the triangles ADC and BEC.

They have two congruent sides that include congruent angles.

Indeed, AC = BC by the condition, because the triangle ABC is isosceles.

Since the lateral sides AC and BC are of equal length, their halves EC

and DC are of equal length too: EC = DC.

Finally, the angle ECD is the common angle.

Thus, the triangles ADC and BEC are congruent, in accordance to the

postulate P1 (SAS) (see the lesson Congruence tests for triangles of the

topic Triangles in the section Geometry in this site).

Hence, the medians AD and BE are of equal length as the corresponding sides

of these triangles.

The proof is completed.

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Whats the answer??? Pleaseeee!!!!
kvasek [131]

To find the 20th term in this sequence, we can simply keep on adding the common difference all the way until we get up to the 20th term.

The common difference is the number that we are adding or subtracting to reach the next term in the sequence.

Notice that the difference between 15 and 12 is 3.

In other words, 12 + 3 = 15.

That 3 that we are adding is our common difference.

So we know that our first term is 12.

Now we can continue the sequence.

12 ⇒ <em>1st term</em>

15 ⇒ <em>2nd term</em>

18 ⇒ <em>3rd term</em>

21 ⇒ <em>4th term</em>

24 ⇒ <em>5th term</em>

27 ⇒ <em>6th term</em>

30 ⇒ <em>7th term</em>

33 ⇒ <em>8th term</em>

36 ⇒ <em>9th term</em>

39 ⇒ <em>10th term</em>

42 ⇒ <em>11th term</em>

45 ⇒ <em>12th term</em>

48 ⇒ <em>13th term</em>

51 ⇒ <em>14th term</em>

54 ⇒ <em>15th term</em>

57 ⇒ <em>16th term</em>

60 ⇒ <em>17th term</em>

63 ⇒ <em>18th term</em>

66 ⇒ <em>19th term</em>

<u>69 ⇒ </u><u><em>20th term</em></u>

<u><em></em></u>

This means that the 20th term of this arithemtic sequence is 69.

5 0
3 years ago
Brainlist for correct answer =]
Vesnalui [34]

Answer:

4

Step-by-step explanation:

The mode is the number that appears the most!!!!

Now lets solve :)

2, 2, 3, 3, 4, 4, 4, 4,

There are 2 2s.

There are 2 3s.

There are 4 4s.

The mode is 4!!! :)

Have an amazing day!!

Plese rate and mark brainliest!!

5 0
2 years ago
Read 2 more answers
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