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mr Goodwill [35]
3 years ago
12

Identify the relation that is also a function. A. {(1, 0) (2, 3) (3, -2) (1, 6)} B. {(3, 2) (3, -2) (3, 6) (3, 8)} C. {(-1, 1) (

0, 0) (1, 1) (2, 4)} D. {(3, 4) (2, 3) (3, 6) (2, 0)}
Mathematics
2 answers:
weqwewe [10]3 years ago
5 0

Answer:

C is the function.

Step-by-step explanation:

C is the only relation which is one-to-one. Each value of x is different and maps to a different value of y.

A for example is one-to-many because   1 maps to 0 and 1 maps to 6 so it is not a function.

Arturiano [62]3 years ago
5 0

Answer:

  C.  {(-1, 1) (0, 0) (1, 1) (2, 4)}

Step-by-step explanation:

A relation is <u>not</u> a function if values of the independent variable are repeated.

  A: (1, -) appears twice

  B: (3, -) appears 4 times

  C: a function

  D: (3, -) appears twice; (2, -) appears twice

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Mice21 [21]

Answer:

5x +2y =38

5(6) +2y = 38

30 +2y= 38

2y= 8

y= 4

(6, 4)

5(0) +2y =38

2y= 38

y= 19

(0,19)

5(-2) +2y= 38

-10 +2y =38

2y= 48

y=24

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7 0
3 years ago
Please help!! its due in 9 minutes
Grace [21]

Answer:

Times it by 40. That is the pattern.

3 0
2 years ago
Read 2 more answers
1. ) Consider the function f(x)=5−7x2,−5≤x≤1
Marat540 [252]
1.) The interval of the value of x is from -5 to 1, inclusive. Remember that what is asked is the absolute value, thus the sign does not matter even if you have to subtract x from 5. Thus, the maximum value would be obtained if the x is smaller, which is 1. The minimum value is obtained when x=-5.

Absolute maximum value: x = - 5
f(-5) = ║5 - 7(-5)^2║ = ║-170║=170


Absolute minimum value: x = 1
f(1) = ║5 - 7(1)^2║ = ║-2║= 2

2.) The Mean Value Theorem (MVT) applies to functions that are continuous and differentiable on the closed and open interval of a to b, respectively. Since the function is a quadratic function, MVT can be applied. Then, this means that there is a value of c which is between a and b. This could be determined using this formula according to MVT:

f'(c)= \frac{f(b)-f(a)}{b-a}

The differentiated form would be f'(x) = -2x. Then,

-2c =  \frac{(4- 0^{2} )-(4- (-1)^{2}) }{0--1}=1

c=- \frac{1}{2}

Thus, x = -1, x = -1/2, and x=0 all lie in the function 4-x^2.
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3 years ago
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OverLord2011 [107]

Answer:

X= 29.5, Y=5

Step-by-step explanation:

If im reading the question correctly it tells you Y is 5, and that X is multiplied by two so half of 59 since we already know that Y is 5 is 29.5

8 0
2 years ago
Read 2 more answers
Wat is x in 1/3x - 11 = 5
kirza4 [7]

=48

Multiply by 1

Add 1 1 11 11

to both sides of the equation

6 0
3 years ago
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