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kirill115 [55]
3 years ago
12

5/10 points, 1 question ( no rude answers )

Mathematics
2 answers:
Vanyuwa [196]3 years ago
5 0
Pretty sure the right answer is the second one
bagirrra123 [75]3 years ago
3 0
The answer is b because all the points lie on the graph
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krek1111 [17]

Answer:

3

Step-by-step explanation:

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Can someone find the slopes of these two please?
GenaCL600 [577]

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you put the points into the slope formula and solve : (y2-y1)/(x2-x1)

1. (4 -1) / (-2 -3)

3 / -5

2. (-1 -(-2)) / (5 - 0)

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Step-by-step explanation:

i hope this helped :)

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Help me please, thank you! :)
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It's 16 units. Just look at the lengths and how many times they are on a shape. The shape is a rectangle, which means there will be two of each length, so just add them all together. 6+6+2+2=16

6 0
3 years ago
The diagram shows rays of light from a point source, P, that reach a detector, D, along two paths. One path is a straight line,
tatiyna
This can solved using the cosine law which is:
c² = a² + b² - 2ab cos θ
Using the values given from the problem
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The answer is the 3rd option.
8 0
2 years ago
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For each part, give a relation that satisfies the condition. a. Reflexive and symmetric but not transitive b. Reflexive and tran
Vesnalui [34]

Answer:

For the set X = {a, b, c}, the following three relations satisfy the required conditions in (a), (b) and (c) respectively.

(a) R = {(a,a), (b,b), (c, c), (a, b), (b, a), (b, c), (c, b)} is reflexive and symmetric but not transitive .

(b) R = {(a, a), (b, b), (c, c), (a, b)} is reflexive and transitive but not symmetric .

(c) R = {(a,a), (a, b), (b, a)} is symmetric and transitive but not reflexive .

Step-by-step explanation:

Before, we go on to check these relations for the desired properties, let us define what it means for a relation to be reflexive, symmetric or transitive.

Given a relation R on a set X,

R is said to be reflexive if for every a \in X, (a,a) \in R.

R is said to be symmetric if for every (a, b) \in R, (b, a) \in R.

R is said to be transitive if (a, b) \in R and (b, c) \in R, then (a, c) \in R.

(a) Let R = {(a,a), (b,b), (c, c), (a, b), (b, a), (b, c), (c, b)}.

Reflexive: (a, a), (b, b), (c, c) \in R

Therefore, R is reflexive.

Symmetric: (a, b) \in R \implies (b, a) \in R

Therefore R is symmetric.

Transitive: (a, b) \in R \ and \ (b, c) \in R but but (a,c) is not in  R.

Therefore, R is not transitive.

Therefore, R is reflexive and symmetric but not transitive .

(b) R = {(a, a), (b, b), (c, c), (a, b)}

Reflexive: (a, a), (b, b) \ and \ (c, c) \in R

Therefore, R is reflexive.

Symmetric: (a, b) \in R \ but \ (b, a) \not \in R

Therefore R is not symmetric.

Transitive: (a, a), (a, b) \in R and (a, b) \in R.

Therefore, R is transitive.

Therefore, R is reflexive and transitive but not symmetric .

(c) R = {(a,a), (a, b), (b, a)}

Reflexive: (a, a) \in R but (b, b) and (c, c) are not in R

R must contain all ordered pairs of the form (x, x) for all x in R to be considered reflexive.

Therefore, R is not reflexive.

Symmetric: (a, b) \in R and (b, a) \in R

Therefore R is symmetric.

Transitive: (a, a), (a, b) \in R and (a, b) \in R.

Therefore, R is transitive.

Therefore, R is symmetric and transitive but not reflexive .

4 0
3 years ago
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