Which relations are functions? Select all that apply.

Mathematics

1 answer:

kaheart [24]3 years ago

7 0

Answer:

A and C

Step-by-step explanation:

B has a domain connected to many other ranges which makes a relationship but confuses the whole function making it incorrect.

For answer D, the x coordinates 1 and 3 both have two y coordinates. That again makes a relationship but confuses the function making that answer incorrect.

In answer A, no domain is repeated in any of the points making it a function.

In answer C, again no domains are paired with more than one range making it a function. That why A and C are the answers

You might be interested in

9 - 3 divided by 1/3 + 1

jolli1 [7]

Answer:

its just 9

Step-by-step explanation:

Divide: 3 : 1 / 3 = 3 / 1 · 3 / 1 = 3 · 3 / 1 · 1 = 9 / 1 = 9 Dividing two fractions is the same as multiplying the first fraction by the reciprocal value of the second fraction. The first sub-step is to find the reciprocal (reverse the numerator and denominator,

6 0

2 years ago

Read 2 more answers

Find a and b if the point p(6,0) and Q(3,2) lie on the graph of ax+ by=12

tiny-mole [99]

to get the equation of any straight line we only need two points off of it, hmmm let's use P and Q here and then let's set the equation in standard form, that is

standard form for a linear equation means

• all coefficients must be integers, no fractions

• only the constant on the right-hand-side

• all variables on the left-hand-side, sorted

• "x" must not have a negative coefficient

$(\stackrel{x_1}{6}~,~\stackrel{y_1}{0})\qquad (\stackrel{x_2}{3}~,~\stackrel{y_2}{2}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{2}-\stackrel{y1}{0}}}{\underset{run} {\underset{x_2}{3}-\underset{x_1}{6}}}\implies \cfrac{2}{-3}\implies -\cfrac{2}{3} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{0}=\stackrel{m}{-\cfrac{2}{3}}(x-\stackrel{x_1}{6})$

$\stackrel{\textit{multiplying both sides by }\stackrel{LCD}{3}}{3(y-0)~~ = ~~3\left( -\cfrac{2}{3}(x-6) \right)}\implies 3y=-2(x-6) \\\\\\ 3y=-2x+12 \implies \stackrel{a}{2} x+\stackrel{b}{3} y=12$