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gregori [183]
3 years ago
15

Is Y=2x-5 linear on nonlinear and why?

Mathematics
2 answers:
sasho [114]3 years ago
7 0

The equation is linear because it is in the form y = mx+b, which is slope intercept form

m = 2 is the slope

b = -5 is the y intercept

Graphing y = 2x-5 produces a straight line.

Vedmedyk [2.9K]3 years ago
6 0
Y=2x-5 is a linear because if you graph it you will get a straight line and it is y=my+b
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Amanda [17]

Answer:

Step-by-step explanation:

rate of interest=(interest×100)÷(principal×time)

=(490×100)÷(7000×2)=3.5 %

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3 years ago
Thirteen less than eight times a number is three more than four times the number.
Lisa [10]

Answer:

Step-by-step explanation:

Let the number = x

8x - 13 = 4x + 3              Add 13

8x = 4x + 3 + 13             Combine

8x = 4x + 16                   Subtract 4x

8x - 4x = 16                    Combine

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x = 16/4

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3 years ago
Express the following relations in the set builder notation. Then, determine whether it is reflexive, symmetric, transitive. Ple
pshichka [43]

Answer:

a)Reflexive, not symmetric, transitive

b)Reflexive, not symmetric, transitive

c)Not reflexive, symmetric, not transitive

d)Reflexive, not symmetric, transitive

Step-by-step explanation:

a)

R=\left \{ (a,b)\epsilon  \mathbb{R} \times \mathbb{R} \mid a \leq b\right \}

The relation R is reflexive for

a\leq a for every real number a

it is not symmetric because 0 is less than 1, but 1 is not less than 0

it is transitive

a\leq and b\leq c\Rightarrow a\leq c

So if aRb and bRc, then aRc

b)  

R=\left \{ (m,n)\epsilon  \mathbb{Z} \times \mathbb{Z} \mid \exists k\in \mathbb{Z} \ni m=kn \right \}

R is reflexive because m=1.m for every integer m

R is not symmetric: 2 is a factor of 4, but 4 is not a factor of 2

R is transitive:  if mRn and nRp if m=kn and n=qp, so m=(kq)p and kq is an integer , so mRp

c)

R=\left \{ (m,n)\epsilon  \mathbb{Z} \times \mathbb{Z} \mid m\neq n\right \}

R is obviously not reflexive because all numbers equals themselves

R is symmetric: if a different to b, then b different to a

R is not transitive: 1R2 and 2R1 (because 1 different to 2), but 1 = 1

d)

R=\left \{ A,B\mid A\subseteq B \right \}

R is reflexive for every set A is a subset of itself

R is not symmetric {1,2} is a subset of {1,2,3} but {1,2,3} is not a subset of {1,2}

R is transitive: if A is subset of B and B is subset of C, then A is subset of C

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lozanna [386]
Answer:
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Step-by-step explanation:

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