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Tanya [424]
2 years ago
10

The congruence theorem that can be used to prove △LON ≅ △LMN is SSS. ASA. SAS. HL.

Mathematics
2 answers:
Zinaida [17]2 years ago
4 0

Answer:

The correct option is SSS (Side-Side-Side) Theorem

Step-by-step explanation:

The question is incomplete because the diagrams of ΔLON and ΔLMN are not given. I have attached the diagram of both triangles below for better understanding of the question.

Consider the diagram attached below. We have to find the congruence theorem which can be used to prove that ΔLON ≅ ΔLMN

We can see in the diagram that both triangle have a common side that is LN. It means 1 side of both triangles is congruent because LN≅LN

Consider the sides ON and MN. Both side have a single bar on them, which means that it is given that both of these side are congruent. Hence ON≅MN

Consider the sides LO and LM. Both side have a double bars on them, which means that it is given that both of these side are also congruent. Hence LO≅LM

SSS theorem states that if all sides of the triangles are congruent, then the triangles themselves are also congruent, which is the same case in this question

likoan [24]2 years ago
3 0

Answer:

D. SSS

Step-by-step explanation:

I hope this helps.

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shepuryov [24]

Answer:

x = 4 , y = ⅕

Step-by-step explanation:

2x + 5y = 9 (1)

3x - 5y = 11 (2)

Add (1) and (2)

5x = 20

x = 4

Using (1), find y

2(4) + 5y = 9

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8 0
2 years ago
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A box with a square base and no top is to be made from a square piece of carboard by cutting 7 in. squares from each corner and
mojhsa [17]

Answer:

The size of the cardboard should be  26  in\ by\ 26in

Step-by-step explanation:

The diagram for this question is shown on the first uploaded image

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The length and the Width of the card is  

      L = W = Z

From the question we are told that 7 in squares were cut from each corner

So the length and the width of the box  would be  

              L = W = (Z - 7)

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Mathematically the volume of the box can be represented as

              V  = Length\ of\ bx  * Width\ of\ box * Height\ of\ box

So the Volume would be

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Substituting values

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7 0
2 years ago
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andrezito [222]

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3 years ago
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LUCKY_DIMON [66]

so hmmm seemingly the graphs meet at -2 and +2 and 0, let's check

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\stackrel{f(x)}{2x^3-x^2-5x}~~ - ~~[\stackrel{g(x)}{-x^2+3x}]\implies 2x^3-x^2-5x+x^2-3x \\\\\\ 2x^3-8x\implies 2(x^3-4x)\implies \displaystyle 2\int\limits_{-2}^{0} (x^3-4x)dx \implies 2\left[ \cfrac{x^4}{4}-2x^2 \right]_{-2}^{0}\implies \boxed{8} \\\\[-0.35em] ~\dotfill

\stackrel{g(x)}{-x^2+3x}~~ - ~~[\stackrel{f(x)}{2x^3-x^2-5x}]\implies -x^2+3x-2x^3+x^2+5x \\\\\\ -2x^3+8x\implies 2(-x^3+4x) \\\\\\ \displaystyle 2\int\limits_{0}^{2} (-x^3+4x)dx \implies 2\left[ -\cfrac{x^4}{4}+2x^2 \right]_{0}^{2}\implies \boxed{8} ~\hfill \boxed{\stackrel{\textit{total area}}{8~~ + ~~8~~ = ~~16}}

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2 years ago
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spayn [35]
57 = 7 + 2x
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Hope this helps a little!!
7 0
3 years ago
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