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3 years ago

Which sets of values belong to the domain and range of a relation?

Mathematics

2 answers:

Nimfa-mama [501]3 years ago

4 0

Answer:

Domain: input values, independent variables

Range: output vales, dependent variables

Step-by-step explanation:

Think of it like a graph: the domain are the x-values and the range is the y-values. if you're doing a problem with time, the time will go on the x-axis and cannot be influenced by the y-values, but the y-vales are depending on what the x-values are (independent/dependent). for the input/output, usually when solving equations on a graph, you plug in the x-value and find the y-value. you're INPUTTING the x-value to receive the OUPUT.

Vikentia [17]3 years ago

3 0

Domain = set of allowed inputs

The input x is the independent variable as it can do whatever it wants without relying on y.

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Range = set of possible outputs

The output is the dependent variable. It depends on what the input x is. Often, we make y the output dependent variable.

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For example, with y = 2x+5, we can plug in anything we want for x (it doesn't need to look to y for guidance or anything). Once we pick something for x, it will directly determine what y is.

Let's say we picked x = 10. That would mean y = 2x+5 = 2*10+5 = 25. The input x = 10 in the domain leads to y = 25 in the range. We see that the output y = 25 depends entirely on the independent input x = 10.

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From 12,000 graphing calculators produced by a manufacturer, an inspector selects a random sample of 550 calculators and finds

maria [59]

Answer:

131

Step-by-step explanation:

Using proportion,

If Out of 550 calculators = 6 are defective,

Then out of 12000 calculators = ? are defective

$\frac{12000 calculators}{550calculators} * 6$

= 21.8 * 6

= 130.8 ≈ 131

∴ Approximately 131 calculators are defective

3 0

3 years ago

A box designer has been charged with the task of determining the surface area of various open boxes (no lid) that can be constru

Viktor [21]

Answer:

1) $S = 2\cdot w\cdot l - 8\cdot x^{2}$ , 2) The domain of S is $0 \leq x \leq \frac{\sqrt{w\cdot l}}{2}$ . The range of S is $0 \leq S \leq 2\cdot w \cdot l$ , 3) $S = 176\,in^{2}$ , 4) $x \approx 4.528\,in$ , 5) $S = 164.830\,in^{2}$

Step-by-step explanation:

1) The function of the box is:

$S = 2\cdot (w - 2\cdot x)\cdot x + 2\cdot (l-2\cdot x)\cdot x +(w-2\cdot x)\cdot (l-2\cdot x)$

$S = 2\cdot w\cdot x - 4\cdot x^{2} + 2\cdot l\cdot x - 4\cdot x^{2} + w\cdot l -2\cdot (l + w)\cdot x + l\cdot w$

$S = 2\cdot (w+l)\cdot x - 8\cdpt x^{2} + 2\cdot w \cdot l - 2\cdot (l+w)\cdot x$

$S = 2\cdot w\cdot l - 8\cdot x^{2}$

2) The maximum cutout is:

$2\cdot w \cdot l - 8\cdot x^{2} = 0$

$w\cdot l - 4\cdot x^{2} = 0$

$4\cdot x^{2} = w\cdot l$

$x = \frac{\sqrt{w\cdot l}}{2}$

The domain of S is $0 \leq x \leq \frac{\sqrt{w\cdot l}}{2}$ . The range of S is $0 \leq S \leq 2\cdot w \cdot l$

3) The surface area when a 1'' x 1'' square is cut out is:

$S = 2\cdot (8\,in)\cdot (11.5\,in)-8\cdot (1\,in)^{2}$

$S = 176\,in^{2}$

4) The size is found by solving the following second-order polynomial:

$20\,in^{2} = 2 \cdot (8\,in)\cdot (11.5\,in)-8\cdot x^{2}$

$20\,in^{2} = 184\,in^{2} - 8\cdot x^{2}$

$8\cdot x^{2} - 164\,in^{2} = 0$

$x \approx 4.528\,in$

5) The equation of the box volume is:

$V = (w-2\cdot x)\cdot (l-2\cdot x) \cdot x$

$V = [w\cdot l -2\cdot (w+l)\cdot x + 4\cdot x^{2}]\cdot x$

$V = w\cdot l \cdot x - 2\cdot (w+l)\cdot x^{2} + 4\cdot x^{3}$

$V = (8\,in)\cdot (11.5\,in)\cdot x - 2\cdot (19.5\,in)\cdot x^{2} + 4\cdot x^{3}$

$V = (92\,in^{2})\cdot x - (39\,in)\cdot x^{2} + 4\cdot x^{3}$

The first derivative of the function is:

$V' = 92\,in^{2} - (78\,in)\cdot x + 12\cdot x^{2}$

The critical points are determined by equalizing the derivative to zero:

$12\cdot x^{2}-(78\,in)\cdot x + 92\,in^{2} = 0$

$x_{1} \approx 4.952\,in$

$x_{2}\approx 1.548\,in$

The second derivative is found afterwards:

$V'' = 24\cdot x - 78\,in$

After evaluating each critical point, it follows that $x_{1}$ is an absolute minimum and $x_{2}$ is an absolute maximum. Hence, the value of the cutoff so that volume is maximized is:

$x \approx 1.548\,in$

The surface area of the box is:

$S = 2\cdot (8\,in)\cdot (11.5\,in)-8\cdot (1.548\,in)^{2}$

$S = 164.830\,in^{2}$

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What is the x-coordinate of the solution to the system?

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I think it would be easiest to solve for x in this system of equations if you did so by elimination:
Start by finding the least common multiple of 3 and 2 to make the Ys cancel:
4x-6y=-54
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