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-BARSIC- [3]
3 years ago
9

ΔMNO is shown.

Mathematics
1 answer:
JulijaS [17]3 years ago
5 0

Answer:

ABC - AAS

DEF - not enough information

GHI - not enough information

JKL - SAS

Step-by-step explanation:

SAS postulate states that if two sides and the included angle of one triangle are congruent to two sides and the included angle of another triangle, then these two triangles are congruent.

AAS postulate states that if two angles and the non-included side one triangle are congruent to two angles and the non-included side of another triangle, then these two triangles are congruent.

HL postulate states that if the hypotenuse and leg of one right triangle are congruent to the hypotenuse and leg of another right triangle, then the two triangles are congruent.

ASA postulate states that if two angles and the included side of one triangle are congruent to two angles and the included side of another triangle, then the triangles are congruent.

SSS postulate states that if three sides of one triangle are congruent to three sides of another triangle, then the triangles are congruent.

1. In triangles MNO and ABC, there are two congruent sides and non-included angle - AAS

2. In triangles MNO and DEF, there are two congruent sides  - there is not enough information

3. In triangles MNO and GHI, there are three congruent angles  - there is not enough information

4. In triangles MNO and JKL, there are two congruent sides and included angle - SAS

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Lostsunrise [7]

The domain is \boxed{x \neq n\pi}, where n is an integer.

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The range is <u>all real numbers</u>.

4 0
2 years ago
If S_1=1,S_2=8 and S_n=S_n-1+2S_n-2 whenever n≥2. Show that S_n=3⋅2n−1+2(−1)n for all n≥1.
Snezhnost [94]

You can try to show this by induction:

• According to the given closed form, we have S_1=3\times2^{1-1}+2(-1)^1=3-2=1, which agrees with the initial value <em>S</em>₁ = 1.

• Assume the closed form is correct for all <em>n</em> up to <em>n</em> = <em>k</em>. In particular, we assume

S_{k-1}=3\times2^{(k-1)-1}+2(-1)^{k-1}=3\times2^{k-2}+2(-1)^{k-1}

and

S_k=3\times2^{k-1}+2(-1)^k

We want to then use this assumption to show the closed form is correct for <em>n</em> = <em>k</em> + 1, or

S_{k+1}=3\times2^{(k+1)-1}+2(-1)^{k+1}=3\times2^k+2(-1)^{k+1}

From the given recurrence, we know

S_{k+1}=S_k+2S_{k-1}

so that

S_{k+1}=3\times2^{k-1}+2(-1)^k + 2\left(3\times2^{k-2}+2(-1)^{k-1}\right)

S_{k+1}=3\times2^{k-1}+2(-1)^k + 3\times2^{k-1}+4(-1)^{k-1}

S_{k+1}=2\times3\times2^{k-1}+(-1)^k\left(2+4(-1)^{-1}\right)

S_{k+1}=3\times2^k-2(-1)^k

S_{k+1}=3\times2^k+2(-1)(-1)^k

\boxed{S_{k+1}=3\times2^k+2(-1)^{k+1}}

which is what we needed. QED

6 0
3 years ago
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Pie

Answer:

49

Step-by-step explanation:

m^2-14m+c

Take the -14

Divide by 2

-14/2 = -7

Square it

(-7)^2 = 49

Add it to each side

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The value of c is 49

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lisabon 2012 [21]
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Vinvika [58]

Answer:

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Step-by-step explanation:

Firstly, please check attachment to have a picture of the triangle we are solving.

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Mathematically, the tangent of an angle is the ratio of the length of the opposite to the length of the adjacent.

In this question, our opposite is 7 while the adjacent is 24.

Thus Tan T = 7/24

6 0
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