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3 years ago

Match the equation with it's graph -4×-2y=8

Mathematics

1 answer:

gogolik [260]3 years ago

8 0

The equation given is: -4x - 2y = 8
we will reorder the equation to in the form of y = mx + c

-4x - 2y = 8
-4x -8 = 2y
y = -2x -4

Now, we will need to generate some points that belong to the graph in order to be able to draw the line.
To do this , we will assume values for x, substitute with these values in the equation and then get the corresponding y.
Doing this, these points will be generated:
(-5,6)
(-4,4)
(-3,2)
(-2,0)
(-1,-2)
(0,-4)
(1,-6)
(2,-8)
(3,-10)
(4,-12)
(5,-14)
Plotting these order pairs on the graph will generate the line shown in the attached image.
Hope this helps :D

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Find the unit rate for 25 miles in 2 hours.

Svetlanka [38]

Answer:

12.5 miles per hour

Step-by-step explanation:

This is a fraction equal to

25 miles ÷ 2 hours

We want a unit rate where

1 is in the denominator,

so we divide top and bottom by 2

25 miles ÷ 2

2 hours ÷ 2

12.5 miles

1 hour

= 12.5 miles per hour

8 0

3 years ago

On Saturday, the fruit and juice bar was selling 27 glasses of fruit punch an hour. By 7 p.m., they had sold 243 glasses. If the

padilas [110]

Answer: 3 hours

Step-by-step explanation:

the answer this can be solved by the function:

324 = 27x + 243

-243 -243

27x = 81

/27 /27

x = 3 hours past 7.

3 0

3 years ago

The model shown represents which of the following?

-BARSIC- [3]

Answer:

its 3

Step-by-step explanation:

selcet 3 it goes plus 3 then back to neg 4 its the 3rd option

4 0

3 years ago

Read 2 more answers

Suppose that a function f(x) is defined for all real values of x except at xequals=c. can anything be said about the existence

Margaret [11]

we are given that

f(x) is defined for all values of x except at x=c

Limit may or may not exist

case-1:

If there is hole at x=c , then limit exist

case-2:

If there is vertical asymptote at x=c , then limit does not exist

Examples:

case-1:

$\lim_{x \to c} \frac{x^2-cx}{(x-c)}$

We can simplify it

$\lim_{x \to c} \frac{x(x-c)}{(x-c)}$

$=\lim_{x \to c} x$

$=c$

so, we can see that limit exist and it's value defined

case-2:

$\lim_{x \to c} \frac{1}{(x-c)}$

Left limit is

$\lim_{x \to c-} \frac{1}{(x-c)}$

$=-\infty$

Right Limit is

$\lim_{x \to c+} \frac{1}{(x-c)}$

$=+\infty$

so, we can see that left limit is not equal to right limit

so, limit does not exist

4 0

3 years ago

the length of one of the diagonals of a quadrilateral is 10 cm and the particular drawn from the opposite vertices to this diago

loris [4]

Answer:

Therefore the area of the quadrilateral =35 cm²

Step-by-step explanation:

Given, the length of one of diagonal of quadrilateral is 10 cm and perpendicular drawn from the opposite vertices to this diagonal are the length of 2.8 cm and 4.2 cm.

A diagonal divided a quadrilateral into two triangle.

Therefore the area of the quadrilateral

= sum of the area of the triangles

$=(\frac{1}{2}\times 10\times 2.8+ \frac{1}{2}\times 10\times 4.2)$ cm² [ area of a triangle $= \frac{1}{2}\times base\times height$ ]

=35 cm²

3 0

3 years ago