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krek1111 [17]
3 years ago
8

Choose all the quadratic pairs that are equivalent. 1 2 (x + 4)2 = 2 x2 + 8x + 15 = 0 (x − 5)2 = 1 x2 − 10x + 26 = 0 3(x − 1)2 +

5 = 0 3x2 − 6x + 8 =
Mathematics
1 answer:
Artemon [7]3 years ago
3 0

Answer:

Only equation 1 and 2 are equal.

Step-by-step explanation:

2 (x + 4)2 = 2

2( x² + 8x+ 16) = 2               Applying the square formula

2x² + 16x+ 32 = 2

2x² + 16x+ 32 -2= 0

2x² + 16x+ 30 = 0

2( x² + 8x+ 15)= 0           Taking 2 as common

x2 + 8x + 15 = 0------------eq 1

x2 + 8x + 15 = 0-------------eq 2

(x − 5)2 = 1

x²-10x+25= 1             Applying the square formula

x²-10x+25- 1= 0

x²-10x+24= 0-------------eq 3

x2 − 10x + 26 = 0 -------------eq 4

3(x − 1)2 + 5 = 0

3( x²-2x+1)+5= 0               Applying the square formula

3x²-6x+3+5= 0

3x²-6x+ 8= 0-------------eq 5

3x2 − 6x + 8 =1

3x2 − 6x + 8 -1=0

3x2 − 6x + 7 =0-------------eq 6

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8 0
3 years ago
Pls helps fast!!
Aleks [24]

Answer:

Option B is correct.

Step-by-step explanation:

We have given a triangle ABC and EDC please look at the figure

We can see that AE and BD are transversal therefore, ∠EAB=∠AED being alternate interior angles

And ∠ACB=∠DCE are vertically opposite angles hence, equal

So, by AA similarity postulate the above to triangles are similar

ΔABC \sim ΔEDC

Therefore, Option B is correct that is Triangle ABC is similar to triangle EDC , because m∠3 = m∠4 and m∠1 = m∠5

NOTE: m∠3 = m∠4 corresponds to m∠ACB=m∠DCE

And m∠1 = m∠5 corresponds to m∠EAB=m∠AED

3 0
3 years ago
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