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3 years ago

3x²+x=2solve by factoring ?

2 answers:

8 0

$3x^2+x=2\\\\3x^2+x-2=0\\\\3x^2+3x-2x-2=0\\\\3x(x+1)-2(x+1)=0\\\\(x+1)(3x-2)=0\\\\x+1=0\ \ \ \ \ \ \ \ \ or\ \ \ \ \ \ \ \ \ 3x-2=0\\x=-1\ \ \ \ \ \ \ \ \ \ \ \ or\ \ \ \ \ \ \ \ \ 3x=2\\x=-1\ \ \ \ \ \ \ \ \ \ \ \ or\ \ \ \ \ \ \ \ \ x=\frac{2}{3}\\\\$

kondor19780726 [428]3 years ago

7 0

$3x^2+x=2\\
3x^2+x-2=0\\
3x^2+3x-2x-2=0\\
3x(x+1)-2(x+1)=0\\
(3x-2)(x+1)=0\\
x=\frac{2}{3} \vee x=-1$

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That is simplest form
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3 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.

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