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Vera_Pavlovna [14]
3 years ago
5

How do you solve the math equation

Mathematics
1 answer:
GalinKa [24]3 years ago
8 0
First of its y-y1 = m(x-x1)
m is the slope, what it is increasing by. 
y1 is the y intercept of your pair. 
X1 is the x intercept of your coordinate pair, plug them in. If you dont know slope the formula to find it is:
y2-y1/x2-x1
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If x = 5 and y = x + 1 , find (x+y)+1/2<br> Pls quickly , ill mark branliest
elena-s [515]

Answer:

23/2

Step-by-step explanation:

If x = 5,

y = 5 + 1 = 6

(5+6) + 1/2 = 11 + 1/2 = 23/2

7 0
3 years ago
(5,65) (7,71) slope
Studentka2010 [4]

Answer:

3

Step-by-step explanation:

The slope formula is m=y2-y1/x2-x1

y2 is 71

y1 is 65

x2 is 7

x1 is 5

71-65 is 6

7-5 is 2

So the slope looks like this:

6/2

Simplify:

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Hope this helps!

6 0
3 years ago
A train will stop every 3/10 of a mile. Which of the following statements are correct?
Tju [1.3M]

Answer:

B

Step-by-step explanation:

Its fourth stop

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4 0
2 years ago
Evaluate.<br> 2(4 what is the correct answer?
Kruka [31]
The Answer Would Be 16

How I Found The Answer:
2 x 2 x 2 x 2 = 16
3 0
3 years ago
Read 2 more answers
How many equivalence relations are there on the set 1, 2, 3]?
Alex787 [66]

Answer:

We need to find how many number of equivalence relations are on the set {1,2,3}

A relation is an equivalence relation if it is reflexive, transitive and symmetric.

equivalence relation R on {1,2,3}

1.For reflexive, it must contain (1,1),(2,2),(3,3)

2.For transitive, it must satisfy: if (x,y)∈R then (y,x)∈R

3. For symmetric, it must satisfy: if (x,y)∈R,(y,z)∈R then (x,z)∈R

Since (1,1),(2,2),(3,3) must be there is R, (1,2),(2,1),(2,3),(3,2),(1,3),(3,1). By symmetry,

we just need to count the number of ways in which we can use the pairs (1,2),(2,3),(1,3) to construct equivalence relations.

This is because if (1,2) is in the relation then (2,1) must be there in the relation.

the relation will be an equivalence relation if we use none of these pairs (1,2),(2,3),(1,3) . There is only one such relation: {(1,1),(2,2),(3,3)}

we can have three possible equivalence relations:

{(1,1),(2,2),(3,3),(1,2),(2,1)}

{(1,1),(2,2),(3,3),(1,3),(3,1)}

{(1,1),(2,2),(3,3),(2,3),(3,2)}

6 0
3 years ago
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