Understanding Functional Relationships

Mathematics

1 answer:

Elza [17]3 years ago

7 0

Answer:

A horizontal line is an example of a functional relationship.

All functions have an independent variable.

All functions have a dependent variable (At least).

Step-by-step explanation:

By description, a relation is a purpose if each input condition has one and only one output value.

A function has at least a secondary variable and an autonomous variable. For instance, given a function:

Y=X

The secondary variable is "y" and the autonomous variable is "x".

The Field of the role is the collection of potential input values and the Range is the collection of potential output values,

Vertical lines have an irregular slope. Since the value of "x" never varies, an upward line is not an illustration of a functional relationship.

In the matter of Horizontal lines, for any x-value, you get the equivalent y-value. These are described as Constant functions.

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Analyze the diagram below and complete the instructions that follow.

Jlenok [28]

Step-by-step explanation:

$scale \: factor = \frac{7}{28} = \frac{1}{4} \\$

$ratio \:of \:surface \:areas\\=\frac{2[5\times 7 + 7\times 3+3\times 5]}{2[20\times 28+ 28\times 12+12\times 20] }\\\\=\frac{[35 + 21+15]}{[560+ 336+240] }\\\\=\frac{71}{1136}\\\\=\frac{1}{16}\\\\= 1\: :\: 16$

3 0

3 years ago

A tank with a capacity of 1000 L is full of a mixture of water and chlorine with a concentration of 0.02 g of chlorine per liter

faltersainse [42]

At the start, the tank contains

(0.02 g/L) * (1000 L) = 20 g

of chlorine. Let c (t ) denote the amount of chlorine (in grams) in the tank at time t .

Pure water is pumped into the tank, so no chlorine is flowing into it, but is flowing out at a rate of

(c (t )/(1000 + (10 - 25)t ) g/L) * (25 L/s) = 5c (t ) /(200 - 3t ) g/s

In case it's unclear why this is the case:

The amount of liquid in the tank at the start is 1000 L. If water is pumped in at a rate of 10 L/s, then after t s there will be (1000 + 10t ) L of liquid in the tank. But we're also removing 25 L from the tank per second, so there is a net "gain" of 10 - 25 = -15 L of liquid each second. So the volume of liquid in the tank at time t is (1000 - 15t ) L. Then the concentration of chlorine per unit volume is c (t ) divided by this volume.

So the amount of chlorine in the tank changes according to

$\dfrac{\mathrm dc(t)}{\mathrm dt}=-\dfrac{5c(t)}{200-3t}$

which is a linear equation. Move the non-derivative term to the left, then multiply both sides by the integrating factor 1/(200 - 5t )^(5/3), then integrate both sides to solve for c (t ):

$\dfrac{\mathrm dc(t)}{\mathrm dt}+\dfrac{5c(t)}{200-3t}=0$