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olasank [31]
3 years ago
14

for two functions, a(x) and b(x), a statement is made that a(x) = b(x) at x=2. what is definitely true about x=2

Mathematics
2 answers:
Musya8 [376]3 years ago
6 0

Answer:

Functions a(x) and b(x) intersect each other at x = 2.

pashok25 [27]3 years ago
3 0

Answer:

The answer is Both a(x) and b(x) have the same output value at x = 2.

Step-by-step explanation:

The answer is Both a(x) and b(x) have the same output value at x = 2. Because when a(x)=b(x) the lines intersect at that point lines intersect they have a point in common. Also, a(x)=b(x) means that the outcomes are the same

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First of all we know the Absolute Value Function that is:

\left | x \right |= \left \{ {{x \ \ \ \ x \geq 0} \atop {-x \ \ \ \ x\ \textless \ 0}} \right.

This is called the Parent Function <em>of the Absolute Value Function.</em>

From the equation:

y=\left | x+2 \right |-3

The term:

\left | x+2 \right |

means that the the Parent Function is <em>shifted</em> two units <em>to the left</em>.

On the other hand, the term:

-3

means that the function \left | x+2 \right | is <em>shifted</em> three units <em>downward. </em>So the result is the graph shown below

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3 years ago
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What would the graph of y=x look like? name three ordered pairs that lie on the line.
ololo11 [35]

-- The graph looks like a line that passes through the origin,
and slopes up to the right at a 45-degree angle.

-- Point #1 on the line:
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-- Point #2 on the line:
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-- Point #3 on the line:
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Given the functions f(x)=1x−2+1 and g(x)=1x+5+9 .
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For this problem you would look at the difference between the two equations. Left and right is determined by the -2 and the +5 in the problems, to get from -2 to +5, you would add 7 but (and i dont really know how to explain the reasoning for this) whenever this number is positive, it goes left and whenever it is negative, it goes right. Figuring out if it shifts up or down is the same, look at the numbers +1 and +9 in the equations. If you go from +1 to +9, you would be adding 8, and this makes it shift upward.

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8 0
3 years ago
John, Joe, and James go fishing. At the end of the day, John comes to collect his third of the fish. However, there is one too m
Dmitry [639]

Answer:

The minimum possible initial amount of fish:52

Step-by-step explanation:

Let's start by saying that

x = is the initial number of fishes

John:

When John arrives:

  • he throws away one fish from the bunch

x-1

  • divides the remaining fish into three.

\dfrac{x-1}{3} + \dfrac{x-1}{3} + \dfrac{x-1}{3}

  • takes a third for himself.

\dfrac{x-1}{3} + \dfrac{x-1}{3}

the remaining fish are expressed by the above expression. Let's call it John

\text{John}=\dfrac{x-1}{3} + \dfrac{x-1}{3}

and simplify it!

\text{John}=\dfrac{2x}{3} - \dfrac{2}{3}

When Joe arrives:

  • he throws away one fish from the remaining bunch

\text{John} -1

  • divides the remaining fish into three

\dfrac{\text{John} -1}{3} + \dfrac{\text{John} -1}{3} + \dfrac{\text{John} -1}{3}

  • takes a third for himself.

\dfrac{\text{John} -1}{3}+ \dfrac{\text{John} -1}{3}

the remaining fish are expressed by the above expression. Let's call it Joe

\text{Joe}=\dfrac{\text{John} -1}{3}+ \dfrac{\text{John} -1}{3}

and simiplify it

\text{Joe}=\dfrac{2}{3}(\text{John}-1)

since we've already expressed John in terms of x, we express the above expression in terms of x as well.

\text{Joe}=\dfrac{2}{3}\left(\dfrac{2x}{3} - \dfrac{2}{3}-1\right)

\text{Joe}=\dfrac{4x}{9} - \dfrac{10}{9}

When James arrives:

We're gonna do this one quickly, since its the same process all over again

\text{James}=\dfrac{\text{Joe} -1}{3}+ \dfrac{\text{Joe} -1}{3}

\text{James}=\dfrac{2}{3}\left(\dfrac{4x}{9} - \dfrac{10}{9}-1\right)

\text{James}=\dfrac{8x}{27} - \dfrac{38}{27}

This is the last remaining pile of fish.

We know that no fish was divided, so the remaining number cannot be a decimal number. <u>We also know that this last pile was a multiple of 3 before a third was taken away by James</u>.

Whatever the last remaining pile was (let's say n), a third is taken away by James. the remaining bunch would be \frac{n}{3}+\frac{n}{3}

hence we've expressed the last pile in terms of n as well.  Since the above 'James' equation and this 'n' equation represent the same thing, we can equate them:

\dfrac{n}{3}+\dfrac{n}{3}=\dfrac{8x}{27} - \dfrac{38}{27}

\dfrac{2n}{3}=\dfrac{8x}{27} - \dfrac{38}{27}

L.H.S must be a Whole Number value and this can be found through trial and error. (Just check at which value of n does 2n/3 give a non-decimal value) (We've also established from before that n is a multiple a of 3, so only use values that are in the table of 3, e.g 3,6,9,12,..

at n = 21, we'll see that 2n/3 is a whole number = 14. (and since this is the value of n to give a whole number answer of 2n/3 we can safely say this is the least possible amount remaining in the pile)

14=\dfrac{8x}{27} - \dfrac{38}{27}

by solving this equation we'll have the value of x, which as we established at the start is the number of initial amount of fish!

14=\dfrac{8x}{27} - \dfrac{38}{27}

x=52

This is minimum possible amount of fish before John threw out the first fish

8 0
3 years ago
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