1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Goshia [24]
2 years ago
15

Which of the following is a true statement about functions ?

Mathematics
1 answer:
Alina [70]2 years ago
6 0

Answer:

D. If f and g are functions then;

(f\bullet g)(x) =f(x)\bullet g(x).

Step-by-step explanation:

The first option is not correct because composition of two functions is not commutative.

The second and third choice will also not yield the same result.

Let

f(x)=x^2 and g(x)=2x.

Then;

(f\bullet g)(x)=2x^3 and f(x)\bullet g(x)=2x^3.

The correct answer is option D.

You might be interested in
Solve the equations for all values of x by completing the square x^2+62=-16x
lorasvet [3.4K]

Answer:

x =  \sqrt{2} - 8\\x =  -\sqrt{2} - 8

Step-by-step explanation:

To complete the square, we first have to get our equation into ax^2 + bx = c form.

First we add 16x to both sides:

x^2 + 16x + 62 = 0

And now we subtract 62 from both sides.

x^2 + 16x = -62

We now have to add (\frac{b}{2})^2 to both sides of the equation. b is 16, so this value becomes (16\div2)^2 = 8^2 = 64.

x^2 + 16x + 64 = -62+64

We can now write the left side of the equation as a perfect square. We know that x+8 will be the solution because 8\cdot8=64 and 8+8=16.

(x+8)^2 = -62 + 64

We can now take the square root of both sides.

x+8 = \sqrt{-62+64}\\\\ x+8 = \pm \sqrt{2}

We can now isolate x on one side by subtracting 8 from both sides.

x = \pm\sqrt{2} - 8

So our solutions are

x =  \sqrt{2} - 8\\x =  -\sqrt{2} - 8

Hope this helped!

4 0
3 years ago
Whats 1-2x+3y for x=-2 and y=-5 ?​
lawyer [7]

Answer:

-10

Step-by-step explanation:

2*-2=-4

3*-5=-15

1-(-4)+(-15)=-10

5 0
3 years ago
Directions: Use a proportion to solve the problem.
spin [16.1K]
So, since she weights 6 times more on earth, say for every lb on the moon  is 6lbs on earth then.

now, if 1lb on the moon is 6lbs on earth, how much is 90 earth lbs on the moon?

\bf \begin{array}{ccll}
moon&earth\\
\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\
1&6\\
m&90
\end{array}\implies \cfrac{1}{m}=\cfrac{6}{90}\implies \cfrac{1\cdot 90}{6}=m
5 0
3 years ago
Help me please.
mart [117]

Answer:

∧ABC ≈ ∧PQR  Angle-Angle Similarity Postulate

∧  ABC ≈ ∧  DEF   Side-Side-Side Similarity Postulate

Step-by-step explanation:

5 0
2 years ago
Find the following sums. Please help.
34kurt

Answer:

5m-n-4p

4a^2+6x-3

Step-by-step explanation:

3m-4n +7p

-5m +9n -6p

+7m -6n -5p

----------------------

Combine like terms

3m-4n +7p

-5m +9n -6p

+7m -6n -5p

----------------------

(3-5+7)m  +(-4+9-6)n  +(7-6-5)p

5m-n-4p

a^2  -3x +1

a^2   +9x -6

2a^2 +0x +2

----------------------

Combine like terms

a^2  -3x +1

a^2   +9x -6

2a^2 +0x +2

----------------------

(1+1+2)a^2 +(-3+9+0)x +(1-6+2)

4a^2+6x-3

6 0
3 years ago
Other questions:
  • Triangle JKL was translated 12 units vertically and then reflected over the y-axis to create triangle J'K'L' What is the value o
    13·1 answer
  • (x²-y²)³+(y²-z²)³+(z²-x²)³/(x-y)³+(y-z)³+(z-x)³ evaluate
    10·1 answer
  • PLEASE HELP!!!! I never learned how to do this...
    13·1 answer
  • Need help, please<br> Find EF
    14·1 answer
  • Find the measure of 2.
    10·2 answers
  • There is a bag filled with 3 blue and 4 red marbles a marble is taken at random from the bag the colour is noted and then it's r
    7·1 answer
  • Can you help me with my pop quiz?​
    6·2 answers
  • REAL ANSWERS TROLL ANSWERS GET REPORTED Create your own graphic organizer that compares and contrasts the Industrial Revolution
    12·1 answer
  • Which is bigger 5t 2 in or 61 in?<br><br> correct=brainliest
    5·2 answers
  • Pls help me thank you
    6·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!