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borishaifa [10]
3 years ago
13

Yes or no, are these functions

Mathematics
1 answer:
Hoochie [10]3 years ago
8 0
Yes it is a function
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Is the following relation a function? (4,2), (1,1), (0,0), (1, -1), (4, -2). Give the domain and range.
Vlada [557]
(x,y)
to be  a function
for every x, ther must be only 1 y to corespond to it
basically, x must NEVER repeat with a different y
list them

ah, we see (1,1) an d(1,-1)
also (4,2) and (4,-2)
double offense
1 repeats with 1 and -1
4 repeats with 2 and -2

not a function

domain is all x values
range is all y values
if they repeat, don't list them

domain=(0,1,4)
range=(-2,-1,0,1,2)
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3 years ago
Which of the following functions is equivalent to the function below? Select two that apply.
Aliun [14]
Can sum1 answer this question cause I also need help on this
7 0
3 years ago
Nico owns 11 instructional piano books. Two are beginner books, six are intermediate books, and three are advanced books. If two
Leona [35]

The probability that he first chooses an advanced book and then chooses a beginner book is 6/121.

<h3>Probability</h3>

Given:

Beginner book=2

Intermediate book=6

Instructional piano books=11

Advanced books=3

Hence:

Probability=(2×3)÷(11×11)

Probability=6/121

Therefore the probability that he first chooses an advanced book and then chooses a beginner book is 6/121.

Learn more about probability here: brainly.com/question/24756209

#SPJ2

7 0
2 years ago
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(PLEASE HELP) Evaluate the expression 2(8 − 4)^2 − 10 ÷ 2.
SOVA2 [1]
To answer your question (PLEASE HELP) Evaluate the expression 2(8 − 4)^2 − 10 ÷ 2.


The answer would be B:27
3 0
3 years ago
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Mhanifa please help! Explain how you can use the perpendicular bisector of a segment to draw an isosceles triangle.
Anna007 [38]

Answer:

If you construct the perpendicular bisector of a line segment, every point on the perpendicular bisector will be the same distance from both point A and point B.

To construct an isosceles triangle, you can start from any point on the perpendicular bisector and draw line segments to point A and to point B

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