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Deffense [45]
2 years ago
15

Write an equation of the line that passes through the points (1,4), (3,7)​

Mathematics
1 answer:
victus00 [196]2 years ago
4 0

Answer:

y=1.5x+2.5

Step-by-step explanation:

First you find the slope:

\frac{7-4}{3-1} or \frac{3}{2}

lets change it into 1.5

Then write it in point-slope: (I used (1,4))

y-4=1.5(x-1)

y-4=1.5x-1.5

y=1.5x+2.5

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Step-by-step explanation:

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Find the equation of the line passing through the points (3,-2) and (3, 4).The answer is x = 3. I'm just wondering how my textbo
Leviafan [203]

Given two points. we can find the equation of a line passing through the points

The formula to be used is:

\frac{y_2-y_1}{x_2-x_!}=\frac{y-y_1}{x-x_!}

where

x_1=3,y_!=-2,x_2=3,y_2=4\frac{4-(-2)}{3-3}=\frac{y-(-2)}{x-3}

=>

\frac{6}{0}=\frac{y+2}{x-3}

The next step is to cross multiply

6(x-3)=0(y+2)

=>

6(x-3)=0

Divide both sides by 6 and make x the subject

x=3

3 0
9 months ago
Factor completely 3a4y3 − 12a3y2 + 6a2y.
Firlakuza [10]

Step-by-step explanation:

3a^4y^3-12a^3y^2+6a^2y=(3a^2y)(a^2y^2)-(3a^2y)(4ay)+(3a^2y)(2)\\\\=(3a^2y)(a^2y^2-4ay+2)

5 0
2 years ago
Could the inverse of a non-function be a function? Explain or give an example.
Kitty [74]

Answer:

The inverse of a non-function mapping is not necessarily a function.

For example, the inverse of the non-function mapping \lbrace (0,\, 0),\, (0,\, 1),\, (1,\, 0),\, (1,\, 1) \rbrace\! is the same as itself (and thus isn't a function, either.)

Step-by-step explanation:

A mapping is a set of pairs of the form (a,\, b). The first entry of each pair is the value of the input. The second entry of the pair would be the value of the output.  

A mapping is a function if and only if for each possible input value x, at most one of the distinct pairs includes x\! as the value of first entry.

For example, the mapping \lbrace (0,\, 0),\, (1,\, 0) \rbrace is a function. However, the mapping \lbrace (0,\, 0),\, (1,\, 0),\, (1,\, 1) \rbrace isn't a function since more than one of the distinct pairs in this mapping include 1 as the value of the first entry.

The inverse of a mapping is obtained by interchanging the two entries of each of the pairs. For example, the inverse of the mapping \lbrace (a_{1},\, b_{1}),\, (a_{2},\, b_{2})\rbrace is the mapping \lbrace (b_{1},\, a_{1}),\, (b_{2},\, a_{2})\rbrace.

Consider mapping \lbrace (0,\, 0),\, (0,\, 1),\, (1,\, 0),\, (1,\, 1) \rbrace\!. This mapping isn't a function since the input value 0 is the first entry of more than one of the pairs.

Invert \lbrace (0,\, 0),\, (0,\, 1),\, (1,\, 0),\, (1,\, 1) \rbrace\! as follows:

  • (0,\, 0) becomes (0,\, 0).
  • (0,\, 1) becomes (1,\, 0).
  • (1,\, 0) becomes (0,\, 1).
  • (1,\, 1) becomes (1,\, 1).

In other words, the inverse of the mapping \lbrace (0,\, 0),\, (0,\, 1),\, (1,\, 0),\, (1,\, 1) \rbrace\! would be \lbrace (0,\, 0),\, (1,\, 0),\, (0,\, 1),\, (1,\, 1) \rbrace\!, which is the same as the original mapping. (Mappings are sets. There is no order between entries within a mapping.)

Thus, \lbrace (0,\, 0),\, (0,\, 1),\, (1,\, 0),\, (1,\, 1) \rbrace\! is an example of a non-function mapping that is still not a function.

More generally, the inverse of non-trivial ellipses (a class of continuous non-function \mathbb{R} \to \mathbb{R} mappings, including circles) are also non-function mappings.

3 0
2 years ago
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