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Y_Kistochka [10]
3 years ago
12

Does this set of ordered pairs represent a function? Why or why not? {(–1, 5), (0,-3), (2,7), (4,0), (7, 5)}

Mathematics
2 answers:
Kisachek [45]3 years ago
8 0
Yes, it does. Because these points do respect the "law" that for each "x" there is only one "y" that corresponds to it
AVprozaik [17]3 years ago
4 0

Answer:

yes, because every x-value corresponds to exactly one y-value

Step-by-step explanation:


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Can some one help me on this math​
wel

the answer would be 50 because 50 dived by 25 equals 2 so then 50 divided by 10 equals 5 and 50 divided 5 equals 10

5 0
3 years ago
9^2 times 9^-6 exponential form
Scrat [10]

The result of the expression in exponential form is 9^-4

<h3>Product of exponents</h3>

Given the expression below

9^2 \times 9^{-6}

According the product law, since the base are equal, we will add the exponent to have:

9^2 \times 9^{-6}=9^{-6+2}\\9^2 \times 9^{-6} = 9^{-4}

Hence the result of the expression in exponential form is 9^-4

Learn more on exponents here; brainly.com/question/11975096

#SPJ1

7 0
2 years ago
What is 6.7 in word form
Gnom [1K]
Answer : Six Point Seven
3 0
4 years ago
Read 2 more answers
Jane bought a jacket that costs $49.95 plus a sales tax of 8%. How much sales tax did Jane pay for the jacket?
DIA [1.3K]

Answer:

$3.996

Step-by-step explanation:

8%=0.08

49.95*0.08=3.996

5 0
3 years ago
PLEASE HELP I HAVE AN HOUR LEFT!!
Yuki888 [10]

The statement that correctly describes the horizontal asymptote of g(x) is:

Limit of g (x) as x approaches plus-or-minus infinity = 6, so g(x) has an asymptote at y = 6.

<h3>What are the asymptotes of a function f(x)?</h3>

  • The vertical asymptotes are the values of x which are outside the domain, which in a fraction are the zeroes of the denominator.
  • The horizontal asymptote is the limit of f(x) as x goes to infinity, as long as this value is different of infinity.

In this problem, the function is:

g(x) = \frac{42x^3 - 15}{7x^3 - 4x^2 - 3}

The horizontal asymptote is given as follows:

y = \lim_{x \rightarrow \infty} g(x) = \lim_{x \rightarrow \infty} \frac{42x^3 - 15}{7x^3 - 4x^2 - 3} = \lim_{x \rightarrow \infty} \frac{42x^3}{7x^3} = \lim_{x \rightarrow \infty} 6 = 6

Hence the correct statement is:

Limit of g (x) as x approaches plus-or-minus infinity = 6, so g(x) has an asymptote at y = 6.

More can be learned about asymptotes at brainly.com/question/16948935

#SPJ1

5 0
2 years ago
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