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trasher [3.6K]
3 years ago
12

If JKL= MNP, which statement must be true

Mathematics
2 answers:
Aleks04 [339]3 years ago
8 0
B is correct answer.
ddd [48]3 years ago
3 0

Answer: B. ∠K ≅ ∠N

Step-by-step explanation:

We know that if two triangles are congruent , then the corresponding angles and sides are congruent by CPCTC.

Given: \triangle {JKL}\cong\triangle{MNP}

Therefore , the corresponding angles and sides of \triangle {JKL}\text{ and }\triangle{MNP} are congruent.

∠J corresponds ∠M

∠K corresponds ∠N

∠L corresponds ∠P

⇒ ∠J ≅ ∠M

∠K ≅ ∠N

∠L ≅ ∠P

\overline{JK}\cong\overline{MN}\\\overline{KL}\cong\overline{NP}\\\overlien{JL}\cong\overline{MP}

From the given options, the statement which must be true :-

∠K ≅ ∠N

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What’s the answer plz help ?
guajiro [1.7K]

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Step-by-step explanation:

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using 3.14 like the problem asks could effect what you're rounding to but either way pi(6)^2=113.04 *4= 452.16

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Determine whether the ffunction X/(x^2-1) is odd even or neither
djyliett [7]

equation is:

f(x) = \frac{x}{ {x}^{2} - 1 }

so if its even f(-x) should be equal to f(x)

that means

f( - x) =   \frac{ - x}{  {( - x)}^{2}  - 1}  \\  \\ f( - x) =     - \frac{ x}{ {x}^{2}  - 1}

so f(x) is not equal to f(-x) so its not even

If its odd than f(-x) should be equal to -f(x)

- f(x) =  - ( \frac{ x}{ { x}^{2} - 1 } ) \\  - f(x) =    - \frac{ x}{ {x}^{2} - 1 }

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Ans:function is odd and not even

3 0
3 years ago
Yes im playing prodigy LOL
Ilya [14]

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0,05

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1)First, Let’s find the mean of the data table

12 + 10 + 12 + 6 + 8 + 4 + 2 + 12 = 66

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2)12 - 8,25 = 3,75

10 - 8,25 = 1,25

12 - 8,25 = 3,75

6 - 8,25 = -1,75

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3 years ago
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3 years ago
Surface area???????????
MAVERICK [17]

Answer:

The option B is the correct answer. 79.18 m²

Step-by-step explanation:

Surface area = Base area +  Lateral surface area

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<u>To find the base area</u>

We can consider the base is a hexagon.

A regular hexagon is a combination of 6 equilateral  triangle

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<u>To find the height</u>

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Therefore the option B is the correct answer. 79.18 m²

7 0
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