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Nezavi [6.7K]
3 years ago
5

Find the value of x in the isosceles triangle shown below. Please help! Thank you

Mathematics
2 answers:
valkas [14]3 years ago
5 0

Answer:

x = √40

Step-by-step explanation:

The Pythagorean theorem formula is

a^{2} + b^{2} = c^{2}

step 1) divide 4 by 2 to get 2

2^{2} + b^{2} = c^{2}

step 2) plug 2 and 6 into the formula

2^{2} + 6^{2} = c^{2}

step 3) multiply 2 by 2 to get 4

4 + 6^{2} = c^{2}

step 4) multiply 6 by 6 to get 36

4 + 36 = c^{2}

step 5) add 4 by 36 to get 40

40 = c^{2}

step 6) take the square root of both sides of the equal sign (the √ cancels out the ²)

√40 = <em>c</em>

step 7) now you have the answer x = √40

Zolol [24]3 years ago
4 0

Answer:

B) x = square root of 40

Step-by-step explanation:

See image below and mark brainliest if helpful

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Let S denote a set of elements. S \times S would denote the set of all ordered pairs of elements of S\!.

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A relation R on S\! is a subset of S \times S. For any two elementsa,\, b \in S, a \sim b if and only if the ordered pair (a,\, b) is in R\!.

 

A relation R on set S is an equivalence relation if it satisfies the following:

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The relation R (on S = \lbrace 1,\, 2,\, 3 \rbrace) in this question is indeed reflexive. (1,\, 1), (2,\, 2), and (3,\, 3) (one pair for each element of S) are all elements of R\!.

R isn't symmetric. (2,\, 3) \in R but (3,\, 2) \not \in R (the pairs in \! R are all ordered.) In other words, 3 isn't equivalent to 2 under R\! even though 2 \sim 3.

Neither is R transitive. (3,\, 1) \in R and (1,\, 2) \in R. However, (3,\, 2) \not \in R. In other words, under relation R\!, 3 \sim 1 and 1 \sim 2 does not imply 3 \sim 2.

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