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blsea [12.9K]
3 years ago
13

Complete the proofs using the most appropriate method. may require CPCTC.

Mathematics
1 answer:
n200080 [17]3 years ago
3 0

For the first one, you did good. I will just suggest a couple things.

Statement              Reason

JK ≅ LM                   Given

<JKM ≅ < LMK        Given  (You did both of these steps so well done.)

MK ≅ MK              Reflexive Property (Your angle pair is congruent but isn't one of the interior angle of the triangles you are trying to prove.)

ΔJMK ≅ ΔLKM        SAS

Problem 2: (You also have a lot of great stuff here.)

Statement          Reason

DE ║ FG              Given

DE ≅ FG              Given

<DEF≅<FGH       Given

<EDF≅<GFH       Corresponding Angles (You don't need to know that F is the midpoint but you got corresponding angle pair which is correct.)

ΔEDF≅ΔGFH        ASA

<DFE≅<FHG       CPCTC  

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then you have 2x=4 simplify

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Answer: 0.375

Step-by-step explanation:

Hey there! If you have any questions feel free to leave them in the comments below.

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Kay [80]

Answer:

1) (x + 3)(3x + 2)

2) x= +/-root6 - 1 by 5

Step-by-step explanation:

3x^2 + 11x + 6 = 0 (mid-term break)

using mid-term break

3x^2 + 9x + 2x + 6 = 0

factor out 3x from first pair and +2 from the second pair

3x(x + 3) + 2(x + 3)

factor out x+3

(x + 3)(3x + 2)

5x^2 + 2x = 1 (completing squares)

rearrange the equation

5x^2 + 2x - 1 = 0

divide both sides by 5 to cancel out the 5 of first term

5x^2/5 + 2x/5 - 1/5 = 0/5

x^2 + 2x/5 - 1/5 = 0

rearranging the equation to gain a+b=c form

x^2 + 2x/5 = 1/5

adding (1/5)^2 on both sides

x^2 + 2x/5 + (1/5)^2 = 1/5 + (1/5)^2

(x + 1/5)^2 = 1/5 + 1/25

(x + 1/5)^2 = 5 + 1 by 25

(x + 1/5)^2 = 6/25

taking square root on both sides

root(x + 1/5)^2 = +/- root(6/25)

x + 1/5 = +/- root6 /5

shifting 1/5 on the other side

x = +/- root6 /5 - 1/5

x = +/- root6 - 1 by 5

x = + root6 - 1 by 5 or x= - root6 - 1 by 5

4 0
3 years ago
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