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Kisachek [45]
4 years ago
14

if you double the input of a function and it results in twice the output and if you triple the input and it result in three time

s the output what can be guessed about the function.
Mathematics
2 answers:
bearhunter [10]4 years ago
8 0

A function behaving as you state is a linear function, which means that the output varies directly with the input.

So, there exists some constant k \in \mathbb{R} such that your function is in the form

f(x) = kx

In fact, if you double the input, you have

f(2x) = k\cdot(2x) = 2\cdot (kx) = 2f(x)

And similarly

f(3x) = k\cdot(3x) = 3\cdot (kx) = 3f(x)

Note that this is a guess, it may happen that output doubles (triples) if you double (triple) the input, and still the function is not like y=kx.

For example, the function

f(x) = \dfrac{x^3}{3} - 2x^2 + \dfrac{14}{3}x -2

is such that

f(1) = 1,\ f(2) = 2,\ f(3) = 3

So, it may seems that doubling and tripling the input doubles and triples the output, but this is not true for every input, for example,

f(2) = 2,\ f(4) = 6

And so doubling the input didn't double the output.

Pepsi [2]4 years ago
4 0

Answer:

The function is most likely directly proportional.

More input results in more output.

Step-by-step explanation:

Apex

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Step-by-step explanation:

Well, as you know, the > symbol means that one number has a greater value than another. For example, 2 > 1. The way we can find that out is using subtraction. Take 54,706 - 45,802, and if that number is positive, then the first number is greater than the other.

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Find the equation of the parabola that has zeros of x = –1 and x = 3 and a y-intercept of (0,–9).
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3 years ago
I keep getting confused I don't understand what I keep doing wrong
rosijanka [135]
\bf \begin{cases}
5x+3y+z=-14\\
2x-2y-z=-13\\
3x+y+3z=-2
\end{cases}

so.. let's us eliminate, say "z" first, so for that, let's pick the 1s and 2nd equations there  \bf \begin{array}{llll}
5x+3y+z=-14\\
2x-2y-z=-13\\
----------\\
\boxed{7x+y+0\ =-27}
\end{array}

so, that's our first two-variables resultant

let's pick other two, and again, eliminate the same "z" variable, this time, let's use the 2nd and 3rd equations   \bf \begin{array}{llll}
2x-2y-z=-13&\times 3\implies &6x-6y-3z=-39\\
3x+y+3z=-2\implies &&3x+y+3z\ =-2\\
&&----------\\
&&\boxed{9x-5y+0\ =-41}
\end{array}

now, we have two two-variables equations, let's use them then
and say, we'll eliminate the "y" variable from them

\bf \begin{array}{llll}
7x+y=-27& \times 5\implies &35x+5y=-135\\
9x-5y=-41\implies &&9x-5y=-41\\
&&-------\\
&&44x+0=-176
\end{array}
\\\\\\
x=\cfrac{-176}{44}\implies \boxed{x=-4}
\\\\\\
\textit{now, that we know x = -4, let's plug that in }7x+y=-27
\\\\\\
7(-4)+y=-27\implies y=-27+28\implies \boxed{y=1}

\bf \textit{now, let's plug those two at }5x+3y+z=-14
\\\\\\
5(-4)+3(1)+z=-14\implies -20+3+z=-14\implies z=-14+17
\\\\\\
\boxed{z=3}
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3 years ago
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