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3 years ago

How do you know whether a relation is a function?(1 point)

Mathematics

1 answer:

torisob [31]3 years ago

5 0

Answer:

2. By determining if each value from one set maps to another set such that each element of the domain pairs with exactly one element of the range.

Step-by-step explanation:

A function is a special relationship where each input has a single output.

Answer options, incorrect statements are underlined below:

1. By determining if each value from one set maps to another set such that each element of the domain pairs with exactly two elements of the range.

Incorrect.

2. By determining if each value from one set maps to another set such that each element of the domain pairs with exactly one element of the range.

Correct.

3. By determining if each value from one set maps to another set such that exactly one element of the domain pairs with exactly two elements of the range.

Incorrect.

4. By determining if each value from one set maps to another set such that exactly one element of the domain pairs with exactly one element of the range.

Incorrect

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The following integral requires a preliminary step such as long division or a change of variables before using the method of par

shtirl [24]

Division yields

$\dfrac{x^4+7}{x^3+2x} = x-\dfrac{2x^2-7}{x^3+2x}$

Now for partial fractions: you're looking for constants a, b, and c such that

$\dfrac{2x^2-7}{x(x^2+2)} = \dfrac ax + \dfrac{bx+c}{x^2+2}$

$\implies 2x^2 - 7 = a(x^2+2) + (bx+c)x = (a+b)x^2+cx + 2a$

which gives a + b = 2, c = 0, and 2a = -7, so that a = -7/2 and b = 11/2. Then

$\dfrac{2x^2-7}{x(x^2+2)} = -\dfrac7{2x} + \dfrac{11x}{2(x^2+2)}$

Now, in the integral we get

$\displaystyle\int\frac{x^4+7}{x^3+2x}\,\mathrm dx = \int\left(x+\frac7{2x} - \frac{11x}{2(x^2+2)}\right)\,\mathrm dx$

The first two terms are trivial to integrate. For the third, substitute y = x ² + 2 and dy = 2x dx to get

$\displaystyle \int x\,\mathrm dx + \frac72\int\frac{\mathrm dx}x - \frac{11}4 \int\frac{\mathrm dy}y \\\\ =\displaystyle \frac{x^2}2+\frac72\ln|x|-\frac{11}4\ln|y| + C \\\\ =\displaystyle \boxed{\frac{x^2}2 + \frac72\ln|x| - \frac{11}4 \ln(x^2+2) + C}$

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3 years ago

Gerald recorded the gains and losses for each play on the football field. Gains were marked with a + and losses with a –. Play 1

Vedmedyk [2.9K]

Answer:

Play 3

Step-by-step explanation:

Given:

$\left\begin{array}{ccccc}{Play}&{1}&{2}&{3}&{4}&{Yards}&{+4}&{-6}&{-2}&{+3}\end{array}\right$

Required [Missing from the question]:

Determine the play with the least change in yard?

To do this, we ignore the sign in front of each yard before we analyze.

So, we have:

Play 1 has a change of 4

Play 2 has 6

Play 3 has 2

Play 4 has 3

From the above analysis, play has the least (which is 2)

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3 years ago

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horrorfan [7]

Honestly idk you’ll need a better way to explain this

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4 years ago

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jolli1 [7]

Answer:

1. x= -4

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Step-by-step explanation:

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3 years ago

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svetlana [45]

Answer:

250

Step-by-step explanation:

150/3=50

50*2=100

150+100=250

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3 years ago

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