What is the domain and range of the relation shown at the right

Mathematics

1 answer:

loris [4]4 years ago

3 0

So the domain is the list of x values that are possible on a line. Looking at this graph, since the only x values that are possible on the line are greater than or equal to 0, the domain is x ≥ 0.

Next, the range is the list of y values that are possible on a line. Looking at this graph, since this line extends both directions vertically, the range is all real numbers.

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Please help me asap!!!

Fed [463]

Answer:

YZ = 4

Step-by-step explanation:

△XZW is simliar to △ YZV

Then use ratios to find the missing side

8/x = 12/6

12x = 48

x = 4

Also good luck on IXL that suuuucccckkksss

4 0

3 years ago

What are the roots of -12x2 + 11x − 3 = 0?

Nataly_w [17]

-12x² + 11x - 3 = 0
x = -(11) +/- √((11)² - 4(-12)(-3))
 2(-12)
x = -11 +/- √(121 - 144)
 -24
x = -11 +/- √(-23)
 -24
x = -11 +/- i√(23)
 -24
x = -11 + i√(23) x = -11 - i√(23)
 -24 -24
x = ¹¹/₂₄ - 0.0416√(23) x = ¹¹/₂₄ + 0.0416√(23)

5 0

3 years ago

Can somebody help me

9966 [12]

Plot the intercepts:

y=-1

x=?

to find the x intercept, you dont solve the equation as it is solved in the slope-intercept form. instead, you simply plug in 0 for y:

0=1/4x-4

0=1/4x-4+4

4=1/4x

4/1/4=1/4x

1/4=0.25

x=16

revise:

y=(0, -1)

x=(16,0)

graph the slope for y=-1/2x-1: start from say point (0,-1)

graph the slope for y=1/4x-4: start from say point (16,0) ; you can just ise the intercept (0,-4) since the graph is diluted small...anyways

the points intersect at (4, -3) so thats the solution to the equations

NOTE: when graphing slopes that are negative either the numerator or the denominator can be graphed negative. so I graphed the first equation with the numerator as the positive and the denominator as the negative : rise= 1 ; run: -2. once i got to the end of the provided graph i started from the -1 y-intercept again and graphed the equation using: rise=-1; run=2

*graphing messes me up, too*