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Misha Larkins [42]
2 years ago
9

Which congruence criteria can be used directly from the information about triangles HIJ and MNO to prove IJ¯¯¯¯¯¯≅MN¯¯¯¯¯¯¯¯¯¯ b

y CPCTC? HI¯¯¯¯¯¯¯≅NO¯¯¯¯¯¯¯¯, ∠I≅∠N, ∠H≅∠O (1 point) ASA congruence AAS congruence SAS congruence HL congruence
Mathematics
1 answer:
skelet666 [1.2K]2 years ago
7 0

Answer: \overline{IJ}\cong \overline{MN}

Step-by-step explanation:

Given: In triangle HIJ and triangle MNO we have

\overline{HI}\cong \overline{NO}, \angle{I}\cong \angle{N} ,\angle{H}\cong \angle{O}

here, HI and NN are the included side between  ∠I & ∠H and ∠N and ∠O.

So , by ASA congruence rule,

ΔHIJ ≅ ΔMNO

So by CPCTC  (corresponding parts of the congruent triangles are congruent)

\overline{IJ}\cong \overline{MN}

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10 • 1/10 =1 a) This is an example of the additive inverse property
saul85 [17]

Answer:

C) Multiplicative Inverse Property

Step-by-step explanation:

Because 1/10 is the inverse of 10.

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The area of a rectangle is 70 m², and the length of the rectangle is 11 m less than three times the width. Find the dimensions o
11111nata11111 [884]

Step-by-step explanation:

the area of a rectangle is

length × width

in our case

length × width = 70 m²

length = 3×width - 11

we can use this second equation in the first and get

(3×width - 11) × width = 70

3×width² - 11×width - 70 = 0

the general solution to a quadratic equation is

x = (-b ± sqrt(b² - 4ac))/(2a)

in our case

a = 3

b = -11

c = -70

x = (11 ± sqrt((-11)² - 4×3×-70))/(2×3) =

= (11 ± sqrt(121 + 840))/6 =

= (11 ± sqrt(961))/6

x1 = (11 + 31)/6 = 42/6 = 7

x2 = (11 - 31)/6 = -20/6 = -10/3

the negative solution is not applicable for a length in an object.

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7 0
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Answer:

0.1875

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The probability that the sample mean will be between 25.5 and 27 years

P(between 25.5 and 27) = 0.1875

3 0
2 years ago
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