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NikAS [45]
3 years ago
11

What else would need to be congruent to show that abc = xyz by asa

Mathematics
1 answer:
pochemuha3 years ago
6 0
Since you're working with the ASA postulate, you're looking to show congruence of the angles at either end of a side. You're given side AC and angle A as congruent with their counterparts. Obviously, you also need to show congruence of angle C with its counterpart, angle Z.

selection B is appropriate
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What’s the correct answer for this?
earnstyle [38]

Answer:

6

Explanation:

According to secant-secant theorem,

(PB)(PA)=(PD)(PC)

(7)(12)=(PD)(14)

NOW

84/14 = PD

PD = 6

4 0
2 years ago
A triangle has two sides of length 9 and 14 what is the smallest possible whole number length for the third side
melisa1 [442]

Let the third side = x.

The sum of any two sides must be greater than the third side

14 + 9 > x

  23 > x

14 + x > 9

   x > -5

9 + x > 14

   x > 5

So 5 < x < 23

6 0
2 years ago
Find two consecutive integers whose sum is 73. Which of the following equations could be used to solve the problem?
max2010maxim [7]
2x  + 1 = 73

2x =  72
x = 36  


the 2 integers are 36 and 37

Its  b
8 0
3 years ago
Read 2 more answers
Carpeting costs $9.95 per yard. If Jamie buys 8 yards, how much will it cost him?
Hatshy [7]

Answer:

$79.6

Step-by-step explanation:

9.95 x 8 = 79.6

8 0
3 years ago
Answer this question
Murljashka [212]

Answer:

a = p * q

b = p * s + q * r

c = r * s

Step-by-step explanation:

In the trinomial ax² + bx + c

a is the coefficient of x²

b is the coefficient of x

c is the numerical term

∵ The trinomial is ax² + bx + c

∵ Its factors are (px + r) and (qx + s)

∴ ax² + bx + c = (px + r)(qx + s)

∵ (px + r)(qx + s) = (px)(qx) + (px)(s) + r(qx) + (r)(s)

∴ (px + r)(qx + s) = pqx² + (psx + qrx) + rs

∴ ax² + bx + c = pqx² + (ps + qr)x + rs

→ By comparing the two sides

∵ ax² = pqx² ⇒ divide both sides by x²

∴ a = pq

∵ bx = (ps + qr)x ⇒ Divide both sides by x

∴ b = ps + qr

∴ c = rs

∴ a = p * q

∴ b = p * s + q * r

∴ c = r * s

5 0
2 years ago
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