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morpeh [17]
3 years ago
12

Is the following relation a function? A. Yes B. No

Mathematics
2 answers:
worty [1.4K]3 years ago
6 0
Yes............................
kykrilka [37]3 years ago
4 0

Answer:

Yes, it's a function.

Step-by-step explanation:

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Let a and b be roots of x² - 4x + 2 = 0. find the value of a/b² +b/a²​
erastovalidia [21]

Answer:

\dfrac{a}{b^2}+\dfrac{b}{a^2}=10

Step-by-step explanation:

Given equation:   x^2-4x+2=0

The roots of the given quadratic equation are the values of x when y=0.

To find the roots, use the quadratic formula:

x=\dfrac{-b \pm \sqrt{b^2-4ac} }{2a}\quad\textsf{when }\:ax^2+bx+c=0

Therefore:

a=1, \quad b=-4, \quad c=2

\begin{aligned}\implies x & =\dfrac{-(-4) \pm \sqrt{(-4)^2-4(1)(2)}}{2(1)}\\& =\dfrac{4 \pm \sqrt{8}}{2}\\& =\dfrac{4 \pm 2\sqrt{2}}{2}\\& =2 \pm \sqrt{2}\end{aligned}

\textsf{Let }a=2+\sqrt{2}

\textsf{Let }b=2-\sqrt{2}

Therefore:

\begin{aligned}\implies \dfrac{a}{b^2}+\dfrac{b}{a^2} & = \dfrac{2+\sqrt{2}}{(2-\sqrt{2})^2}+\dfrac{2-\sqrt{2}}{(2+\sqrt{2})^2}\\\\& = \dfrac{2+\sqrt{2}}{6-4\sqrt{2}}+\dfrac{2-\sqrt{2}}{6+4\sqrt{2}}\\\\& = \dfrac{(2+\sqrt{2})(6+4\sqrt{2})+(2-\sqrt{2})(6-4\sqrt{2})}{(6-4\sqrt{2})(6+4\sqrt{2})}\\\\& = \dfrac{12+8\sqrt{2}+6\sqrt{2}+8+12-8\sqrt{2}-6\sqrt{2}+8}{36+24\sqrt{2}-24\sqrt{2}-32}\\\\& = \dfrac{40}{4}\\\\& = 10\end{aligned}

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Given that there are 60 minutes in an hour and Mr. Sneed runs his new machine eight hours per day how many bottles does this mac
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Answer:

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Step-by-step explanation:

so u day 1+1 then u get 11, but is 2 so u say 112 so tysm for letting me be dum, also i just need points. sorry other ppl :) BYEE

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Solve the system of linear equations by graphing x - 3y = 3 4x + 3y = - 3​
Colt1911 [192]

Answer:

Assuming your equations are:

x-3y=3

4x+3y=-3

The answers are x = 0 and y = -1.

See the image for the graph.

If this solution helped you, consider upvoting it and giving a brainliest!

Step-by-step explanation:

See the image for the graph.

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