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Andrei [34K]
3 years ago
7

Which equation does the graph of the systems of equations solve? 2 linear graphs. They intersect at 1,4

Mathematics
1 answer:
Pepsi [2]3 years ago
6 0

Answer:

See below.

Step-by-step explanation:

There is an infinite n umber of systems of equations that has (1, 4) as its solution. Are you given choices? Try x = 1 and y = 4 in each equation of the choices. The set of two equations that are true when those values of x and y are used is the answer.

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Help PLEASE I will give you 30 points please
ohaa [14]

Answer

1) true

2)false

3)false

4)true

5)true

Step-by-step explanation:

7 0
3 years ago
Which of the following equations is equivalent to S=pi r^2 h
lara31 [8.8K]

The equation that is equivalent to S=pi r^2 h is h = S/pir^2

<h3>Subject of formula</h3>

This is a way of representing a variable with another. Given the equation

S=pi r^2 h

We are to make 'h" the subject of the formula.

Divide both sides by pir^2

S/pir^2 =pi r^2 h/pir^2

h = S/pir^2

Hence the equation that is equivalent to S=pi r^2 h is h = S/pir^2

Learn more on subject of formula here: brainly.com/question/657646

#SPJ1

4 0
2 years ago
SYSTEM OF LINEAR EQUATIONS
Effectus [21]

Answer:

Step-by-step explanation:

x-y=11

2x+y=19

---------------

x=y+11

2(y+11)+y=19

2y+22+y=19

3y+22=19

3y=19-22

3y=-3

y=-3/3

y=-1

x-(-1)=11

x+1=11

x=11-1

x=10

Answer: x=10, y=-1. (10, -1).

-----------------------------------------

-6x+5y=1

6x+4y=-10

-------------------

9y=-9

y=-9/9

y=-1

6x+4(-1)=-10

6x-4=-10

6x=-10+4

6x=-6

x=-6/6

x=-1

Answer: x=-1, y=-1. (-1, -1).

------------------------------------

2x+9y=25

-4x-9y=-23

-------------------

-2x=2

x=2/-2

x=-1

2(-1)+9y=25

-2+9y=25

9y=25-(-2)

9y=25+2

9y=27

y=27/9

y=3

Answer: x=-1, y=3. (-1, 3).

4 0
3 years ago
Read 2 more answers
Check whether the relation R on the set S = {1, 2, 3} is an equivalent
kozerog [31]

Answer:

R isn't an equivalence relation. It is reflexive but neither symmetric nor transitive.

Step-by-step explanation:

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

For example, with S = \lbrace 1,\, 2,\, 3 \rbrace, (3,\, 2) and (2,\, 3) are both members of S \times S. However, (3,\, 2) \ne (2,\, 3) because the pairs are ordered.

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:

  • Reflexivity: for any a \in S, the relation R needs to ensure that a \sim a (that is: (a,\, a) \in R.)
  • Symmetry: for any a,\, b \in S, a \sim b if and only if b \sim a. In other words, either both (a,\, b) and (b,\, a) are in R, or neither is in R\!.
  • Transitivity: for any a,\, b,\, c \in S, if a \sim b and b \sim c, then a \sim c. In other words, if (a,\, b) and (b,\, c) are both in R, then (a,\, c) also needs to be in R\!.

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.

3 0
3 years ago
Which equation is equivalent to<br> log5x^3 - logx^2 = 2?
Oksanka [162]

Answer:

the second option and x=20

4 0
3 years ago
Read 2 more answers
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