All categories

3 years ago

Select the relations that are functions.

Mathematics

1 answer:

Arlecino [84]3 years ago

8 0

Answer:

Depending on how the input of each function defined,

The first choice $\left\lbrace(a, 1),\, (6, 1), \, (C, 1)\right\rbrace$ ,
The third choice $\left\lbrace(1, a),\, (2, a), \, (3, a)\right\rbrace$
The fourth choice $\left\lbrace(a, a),\, (b, b), \, (c, c)\right\rbrace$

might be functions.

Step-by-step explanation:

A function between two sets (domain and range) should

be defined for all elements in the domain, and
map each element from the domain to exactly one element in the range.

The second choice can't be a function since the element $a$ from the domain is mapped to more than one element in the range.

Keep in mind that a function should be defined for all elements in its domain. For the first relation to be a function, its domain needs to be $\lbrace a,\, 6, \, C\rbrace$ . Similarly, the domain for the third and fourth relations should be $\lbrace 1,\, 2, \, 3\rbrace$ and $\lbrace a,\, b, \, c\rbrace$

You might be interested in

Please answer the question

maw [93]

The answer would be 208 if you find the area of each quadrilateral and add them together.

6 0

3 years ago

Read 2 more answers

Help asap pls and thx

vazorg [7]

Step-by-step explanation:

from what I can see there are 2 options to make the 2 triangles congruent :

either the opposite sides must be equal or the neighboring sides must be equal.

so, either

5x = 3x + 10

and

5x - 2 = 4x + 3

5x = 4x + 3

and

5x - 2 = 3x + 10

for the first

5x = 3x + 10

2x = 10

x = 5

control

5x - 2 = 4w + 3

5×5 - 2 = 4×5 + 3

23 = 23

correct.

for the second

5x = 4x + 3

x = 3

control

5×3 - 2 = 3×3 + 10

15 - 2 = 9 + 10

13 = 19

no, not a valid solution.

so, x = 5

5 0

2 years ago

Find the Taylor series for f(x) centered at the given value of a. [Assume that f has a power series expansion. Do not show that

juin [17]

Answer:

The Taylor series of f(x) around the point a, can be written as:

$f(x) = f(a) + \frac{df}{dx}(a)*(x -a) + (1/2!)\frac{d^2f}{dx^2}(a)*(x - a)^2 + .....$

Here we have:

f(x) = 4*cos(x)

a = 7*pi

then, let's calculate each part:

f(a) = 4*cos(7*pi) = -4

df/dx = -4*sin(x)

(df/dx)(a) = -4*sin(7*pi) = 0

(d^2f)/(dx^2) = -4*cos(x)

(d^2f)/(dx^2)(a) = -4*cos(7*pi) = 4

Here we already can see two things:

the odd derivatives will have a sin(x) function that is zero when evaluated in x = 7*pi, and we also can see that the sign will alternate between consecutive terms.

so we only will work with the even powers of the series:

f(x) = -4 + (1/2!)*4*(x - 7*pi)^2 - (1/4!)*4*(x - 7*pi)^4 + ....

So we can write it as:

f(x) = ∑fₙ

Such that the n-th term can written as:

$fn = (-1)^{2n + 1}*4*(x - 7*pi)^{2n}$