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goldenfox [79]
4 years ago
9

Which part of the proof does the fourth statement and reason represent?

Mathematics
1 answer:
Neporo4naja [7]4 years ago
7 0

Answer:

Option B is correct.

Reason:It is an argument

Step-by-step explanation:

Given: AB \cong BC  and BC \cong EF

By transitive property:  a = b  and b = c then, a =c

⇒AB \cong EF

AB = EF                                [def of \cong segment]

By Segment Addition Postulate states that given  two points A and C, and a third point B lies on the line segment AC if and only if the distances between the  points satisfy the equation

AB + BC = AC.

Since, the line segment EG , F lies on the line segment;

then, by segment addition postulates we have;

EG = EF + FG

By substitution AB=EF

⇒ EG = AB + FG                     hence proved!

Since, in the fourth statement reason is:  It is an argument

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Part A: Yes there is. As the number of workers increases, so does the production.

Part B:    y = 5x + 2

Part C: The slope indicates that for every person there is 5 units being produced, and the y-intercept indicates that 2 units are being made every day, regardless of if there are no workers ( this, however is unrealistic but it is how the data is )

6 0
2 years ago
Factor. 2xy+5x−12y−30
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Factor. 2xy+5x−12y−30


(2y+5) (x-6)
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3 years ago
Helpppppppppppppppppppppppppppppppppppppppp
polet [3.4K]

Step-by-step explanation:

let : the pair of jeans = X

a. the equation is :

4(X-3) + 8 = 92

b. 4(X-3) = 92-8

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4X = 96

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_________

For the solution x = 2.6, the left side of equation

(2x - 6) => would be 2(2.6)-6 = 5.2-6 = -8 ( the negative value),

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4 0
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There are 6 children at a birthday party. Each child gets 2 1/3 cups of fruit punch. How many cups of fruit punch are needed for
fredd [130]
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3 years ago
Assume u4 is not a linear combination of {u1,u2,u3}
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Answer:

A. {u1,u2,u3,u4} is a linearly independent set of vectors unless one of {u1,u2,u3} is the zero vector.

Step-by-step explanation:

Given that u4 is not a linear combination of {u1,u2,u3}

This means there is no possibility to write u4 = au1+bu2+cu3 for three scalars a,b,and c.

This gives that u_4-au_1-bu_2-cu_3 \neq 0

This implies that these four vectors are not linearly dependent but linearly independent.

Hence option a is right.

A. {u1,u2,u3,u4} is a linearly independent set of vectors unless one of {u1,u2,u3} is the zero vector.

8 0
3 years ago
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