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2 years ago

Which Graph Represents A function ? NEED ANSWER ASAP !!!!

Mathematics

2 answers:

ANEK [815]2 years ago

7 0

Answer:

The very last one is a function

Step-by-step explanation:

The last one is a function because there are no y values that repeat. For example, If the points on one graph are like this; (1, 3) (2, 3); then it cannot be a function, the y values cannot repeat if it is a function.

To make things easier, just try to imagine vertical drawing lines through each point in every graph, if this imaginary vertical line passes more than one point in any graph, then that graph is not a function.

I hope this helps!! :D

Alexeev081 [22]2 years ago

3 0

Answer:

The last graph represents a function.

Explanation:

A function cannot be one-to-many, i.e., one value of x cannot produce multiple values of y.

In the first graph, when x = 1, y = 3 and y= -2, so it is not a function.

In the second graph, when x = 2, y = 2 and y = -2, so it's not a function.

In the third graph, when x = -3, y = 2 and y = -2, so it's not a function.

∴ In all the graphs except the last, single values of x give multiple values of y on the graph, so they are not functions.

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Which fraction set is equivalent to 5/6 ÷ 7?

Ymorist [56]

10/12 = 15/18 = 20/24

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3 years ago

This graph is a function. True or false

pashok25 [27]

Answer:

True, every input has one output

Step-by-step explanation:

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3 years ago

Which ordered pair is a solution to the system of linear equations 1/2x-3/4y=11/60 and 2/5x+1/6y=3/10

natka813 [3]

ANSWER

$( \frac{2}{3} , \frac{1}{5} )$

EXPLANATION

The first equation is

$\frac{1}{2} x - \frac{3}{4} y = \frac{11}{60} ...(1)$

The second equation is

$\frac{2}{5} x + \frac{1}{6} y = \frac{3}{10} ...(2)$

We want to eliminate y, so we multiply the first equation by

$\frac{4}{5}$

$\frac{4}{5} \times \frac{1}{2} x - \frac{4}{5} \times \frac{3}{4} y = \frac{11}{60} \times \frac{4}{5}$

$\frac{2}{5} x - \frac{3}{5} y = \frac{11}{75} ...(3)$

We now subtract equation (3) from (2)

$(\frac{2}{3} x - \frac{2}{3} x )+ ( \frac{1}{6} y - - \frac{3}{5}y ) =( \frac{3}{10} - \frac{11}{75} )$

$\frac{1}{6} y + \frac{3}{5}y =\frac{3}{10} - \frac{11}{75}$

$\frac{23}{30}y = \frac{23}{150}$

Multiply both sides by

$\frac{30}{23}$

$\implies \: \frac{30}{23} \times \frac{23}{30}y= \frac{23}{150} \times \frac{30}{23}$

$\implies \: y = \frac{1}{5}$

Substitute into the first equation to solve for x .

$\frac{1}{2} x - \frac{3}{4} \times \frac{1}{5} = \frac{11}{60}$

Multiply to obtain

$\frac{1}{2} x - \frac{3}{20} = \frac{11}{60}$

$\frac{1}{2} x = \frac{11}{60} + \frac{3}{20}$

$\frac{1}{2} x = \frac{1}{3}$

Multiply both sides by 2.

$2 \times \frac{1}{2} x =2 \times \frac{1}{3}$

$x = \frac{2}{3}$

The solution is

$( \frac{2}{3} , \frac{1}{5} )$

5 0

3 years ago

Read 2 more answers

Segment PE has endpoints P(-4,4) and E(0,-4). Find the coordinates of point Q such that PQ:QE is 1:3. Enter the coordinates belo

Natalka [10]

Answer:

(-3, 2)

Step-by-step explanation:

Given that point Q, partitions segment PE, such that PQ:QE is 1:3, coordinates of point Q is found using the formula below:

$x = \frac{mx_2 + nx_1}{m + n}$

$y = \frac{my_2 + ny_1}{m + n}$

Where,

$P(-4, 4) = (x_1, y_1)$

$E(0, -4) = (x_2, y_2)$

$m = 1, n = 3$

Plug in the necessary values to find x and y coordinates for point Q as follows:

$x = \frac{mx_2 + nx_1}{m + n}$

$x = \frac{1(0) + 3(-4)}{1 + 3}$

$x = \frac{0 - 12}{4}$

$x = \frac{-12}{4}$

$x = -3$

$y = \frac{my_2 + ny_1}{m + n}$

$y = \frac{1(-4) + 3(4)}{1 + 3}$

$y = \frac{-4 + 12}{4}$

$y = \frac{8}{4}$

$y = 2$

The coordinates of the point Q are (-3, 2))

8 0

3 years ago

1 4/7x4/5 in simplest form

hjlf

Answer:

1 9/35

Step-by-step explanation:

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3 years ago