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3 years ago

Is a relation always a function ? is a function always a relation ? explain

1 answer:

3 0

It just so happens that it's always the same y for each x, but it is only that one y. So this is a function; it's just an extremely boring function! ... So this is a relation, but it is not a function.

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Rewrite 2(x + 3y - 2z) using the Destributive Property

zloy xaker [14]

Answer:

2x+6y-4z

Step-by-step explanation:

multiply the two by each term and coefficient inside of the parentheses

6 0

3 years ago

SOMEONE PLEASE HELP ME WITH THIS!!<br> IMAGE DOWN BELOW!!<br> ILL GIVE YOU BRAINLIST ANSWER!!

larisa [96]

Answer:

7.235 cm

Step-by-step explanation:

Pythagorean theory

A^2 + B^2 = C^2

A = 4.7

B = 5.5

4.7^2 + 5.5^2 = C^2

22.09 + 30.25 = C^2

√52.34 = √C^2

7.235 cm ≈ C

3 0

2 years ago

The rabbot population in a certian area is 200% of last years population there are 1100 rabbits this year how many were there la

Dahasolnce [82]

1100/2

=550

because 100% is 500 and 100% is last year's population

6 0

3 years ago

MATH will give brainliest

Zepler [3.9K]

Answer:

4th choice

Step-by-step explanation:

Volume of cylinder: 2(pi)r^2+2pi(r)h

3 0

3 years ago

Read 2 more answers

If f(x) = x-1/3 and g(x)= 3x+1, what is (f o g)(x)?

N76 [4]

Answer:

$(f o g)(x) = 3x + \frac{2}{3}$

Step-by-step explanation:

We have the function $f(x) = x-\frac{1}{3}$ and we have the function $g(x) = 3x + 1$ . We want to find g(x) composed with f(x)

Then, the function (f o g)(x) is the same since f(g(x))

That is, you must do x = g(x) and then enter g(x) into the function f(x).

$f(g(x)) = (g(x)) -\frac{1}{3}$

$f(g(x)) = (3x + 1) -\frac{1}{3}$

Simplifying, we obtain:

$(f o g)(x) = 3x + 1 -\frac{1}{3}\\\\(f o g)(x) = 3x + \frac{2}{3}$

Finally. The composite function is:

$(f o g)(x) = 3x + \frac{2}{3}$

6 0

3 years ago