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Amiraneli [1.4K]
3 years ago
5

Which fraction is not equivalent to 15/20

Mathematics
2 answers:
Ksivusya [100]3 years ago
4 0
Are there answer choices? If not 1/2 and 2/3 are not equivalent to 15/20
alexdok [17]3 years ago
3 0
D.
\frac{15}{20}=\frac{5}{5} * \frac{3}{4}=\frac{3}{4}\\\\ \frac{3}{4}\\ \frac{9}{12}=\frac{3}{3}*\frac{3}{4}=\frac{3}{4}\\ \frac{12}{16} = \frac{4}{4} * \frac{3}{4}=\frac{3}{4}\\ \frac{25}{30}=\frac{5}{5}*\frac{5}{6} \\ \frac{5}{6} is the only fraction ≠ \frac{3}{4}, so D is your answer,
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X-y=5 and x^2y=5x+6​
sergeinik [125]

By applying algebraic handling on the two equations, we find the following three <em>solution</em> pairs: x₁ ≈ 5.693 ,y₁ ≈ 10.693; x₂ ≈ 1.430, y₂ ≈ 6.430; x₃ ≈ - 0.737, y₃ ≈ 4.263.

<h3>How to solve a system of equations</h3>

In this question we have a system formed by a <em>linear</em> equation and a <em>non-linear</em> equation, both with no <em>trascendent</em> elements and whose solution can be found easily by algebraic handling:

x - y = 5      (1)

x² · y = 5 · x + 6       (2)

By (1):

y = x + 5

By substituting on (2):

x² · (x + 5) = 5 · x + 6

x³ + 5 · x² - 5 · x - 6 = 0

(x + 5.693) · (x - 1.430) · (x + 0.737) = 0

There are three solutions: x₁ ≈ 5.693, x₂ ≈ 1.430, x₃ ≈ - 0.737

And the y-values are found by evaluating on (1):

y = x + 5

x₁ ≈ 5.693

y₁ ≈ 10.693

x₂ ≈ 1.430

y₂ ≈ 6.430

x₃ ≈ - 0.737

y₃ ≈ 4.263

By applying algebraic handling on the two equations, we find the following three <em>solution</em> pairs: x₁ ≈ 5.693 ,y₁ ≈ 10.693; x₂ ≈ 1.430, y₂ ≈ 6.430; x₃ ≈ - 0.737, y₃ ≈ 4.263.

To learn more on nonlinear equations: brainly.com/question/20242917

#SPJ1

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2 years ago
Is this a function. ?
goldenfox [79]

Answer:

No

Step-by-step explanation:

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