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
mylen [45]
3 years ago
7

Is the following relation a function?

Mathematics
2 answers:
IRINA_888 [86]3 years ago
8 0

The answer is B (no)

Eduardwww [97]3 years ago
6 0
The answer is No.
Because the 0 in x oval is pointing to BOTH 1 and 3 in y oval.
And that must not happen in order to relation be a function.
The following conditions must be satisfied for relation to be a function:
1. condition: EVERY element in first oval has to be connected to some element in second oval.
2. condition: Elements in first oval MUST NOT have more than one connection with elements in second oval.
You might be interested in
Find the area and circumference of a circle which is 5cm​
Nonamiya [84]

Answer: 31.416

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
0.56 as a fraction in simplest form.
Vlada [557]

Answer:

14/25

Step-by-step explanation:

0.56=56/100=28/50=14/25

4 0
3 years ago
PLZ HELP.
Allushta [10]

Answer: c

Step-by-step explanation:

4 0
3 years ago
If S_1=1,S_2=8 and S_n=S_n-1+2S_n-2 whenever n≥2. Show that S_n=3⋅2n−1+2(−1)n for all n≥1.
Snezhnost [94]

You can try to show this by induction:

• According to the given closed form, we have S_1=3\times2^{1-1}+2(-1)^1=3-2=1, which agrees with the initial value <em>S</em>₁ = 1.

• Assume the closed form is correct for all <em>n</em> up to <em>n</em> = <em>k</em>. In particular, we assume

S_{k-1}=3\times2^{(k-1)-1}+2(-1)^{k-1}=3\times2^{k-2}+2(-1)^{k-1}

and

S_k=3\times2^{k-1}+2(-1)^k

We want to then use this assumption to show the closed form is correct for <em>n</em> = <em>k</em> + 1, or

S_{k+1}=3\times2^{(k+1)-1}+2(-1)^{k+1}=3\times2^k+2(-1)^{k+1}

From the given recurrence, we know

S_{k+1}=S_k+2S_{k-1}

so that

S_{k+1}=3\times2^{k-1}+2(-1)^k + 2\left(3\times2^{k-2}+2(-1)^{k-1}\right)

S_{k+1}=3\times2^{k-1}+2(-1)^k + 3\times2^{k-1}+4(-1)^{k-1}

S_{k+1}=2\times3\times2^{k-1}+(-1)^k\left(2+4(-1)^{-1}\right)

S_{k+1}=3\times2^k-2(-1)^k

S_{k+1}=3\times2^k+2(-1)(-1)^k

\boxed{S_{k+1}=3\times2^k+2(-1)^{k+1}}

which is what we needed. QED

6 0
3 years ago
Melissa put her cat, Herman, on a diet and kept track of his weight. At the end of 4 weeks, she recorded Herman's weight change
vampirchik [111]
-103
if he lost the same weight each week that means -412 (the weight he lost) ÷ 4 (number of weeks) = -103 (amount of weight lost in one week)
8 0
3 years ago
Read 2 more answers
Other questions:
  • Complete the synthetic division problem below. What is the quotient in polynomial form?
    5·2 answers
  • HELP HELP PLEASE ASAP SIMPLE QUESTION!!!! WILL MARK BRAINLIEST AND 5 STARS
    11·1 answer
  • Which are the roots of the quadratic function f(b) = b2 – 75? check all that apply. b = 5 square root of 3 b = -5 square root of
    7·1 answer
  • Is 890,353 mL bigger than 89 L 353mL
    12·1 answer
  • Which statement is true about the terms of the linear expression 13-4x in standard form?
    6·1 answer
  • Question 8 of 10
    6·1 answer
  • What is the eighth term in the following sequence?
    10·2 answers
  • Anyone know how to do this...
    5·1 answer
  • Hhhhhheelllp ASAPpppppp
    6·2 answers
  • Solve the inequality
    5·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!