Which set of ordered pairs represents a function?

aliina [53]

52 minutes 30 seconds Take 60 minutes divide it by 8 to figure out what 1/8 of an hour is. Then multiply our answer by 7 to find out how many minutes are in 7/8 of an hour.

6 0

3 years ago

Read 2 more answers

A small cube has the volume shown. Its side length is 1.5 in less than a second larger cube. What is

eduard

The volume of the larger cube is 227 inches cube.

<h3 /><h3 /><h3>Volume of a cube is describe below:</h3>

v = L³

where

L = length

The small cube has the following volume:

v = 95 in³

Let

The side length of the small cube = x

Therefore,

95 = x³

x = ∛95

x = 4.56290264

x ≈ 4.6

Its side length is 1.5 inches less than a second larger cube. Therefore,

length of larger length cube = (x + 1.5 )inches = 4.6 + 1.5 = 6.1 inches

Therefore, the volume of the larger cube can be found as follows:

Volume = 6.1³

volume = 226.981

volume ≈ 227 inches³

learn more on cubes here:brainly.com/question/2671825?referrer=searchResults

6 0

2 years ago

Write an equation that goes through (8,1) and is perpendicular to 2y + 4x =12

Serggg [28]

Answer:

2y -x + 6 = 0

Step-by-step explanation:

Here a equation and a point is given to us and we are interested in finding a equation which is perpendicular to the given equation and passes through the given point .

The given equation is ,

$\sf \longrightarrow 2y + 4x = 12 \\$

$\sf \longrightarrow 4x + 2y = 12$

Firstly convert this into slope intercept form , to find out the slope of the line.

$\sf \longrightarrow 2y = -4x +12\\$

$\sf \longrightarrow y =\dfrac{-4x+12}{2}\\$

$\sf \longrightarrow y =\dfrac{-4x}{2}+\dfrac{12}{2}\\$

$\sf \longrightarrow \red{ y = -2x + 6 }$

Now on comparing it to slope intercept form which is y = mx + c , we have ,

m = -2

And as we know that the product of slopes of two perpendicular lines is -1 . So the slope of the perpendicular line will be negative reciprocal of the slope of the given line. Therefore ,

$\sf \longrightarrow m_{\perp}=\dfrac{-1}{-2}=\red{\dfrac{1}{2}}$