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AlladinOne [14]
3 years ago
7

State whether the function is bounded above, bounded below, or bounded. y = 2x

Mathematics
1 answer:
kotykmax [81]3 years ago
3 0
Note: Type "2^x" without quotes to indicate 2^x

When you graph this function, you'll see that we have the exponential curve sloping upward as x increases. The function increases upward without bound. In other words, it goes on forever upward. There are no restrictions on the "ceiling" so to speak of this function curve.

There is a floor though. The floor in this case is like an electric fence that the function never touches. It only gets closer. This is known as the "horizontal asymptote". In this case, the horizontal asymptote is the line y = 0 which lays perfectly over the x axis.

So because of this "fence", the function is bounded below. It does not go forever in the negative y direction.

Final Answer: Choice B) Bounded below
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If DE=4x+10, EF =2x-1, and DF=9x-15, find DF
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Step-by-step explanation:

If DE = 4x+10,EF =2X -1, and DF= 9x - 15 find DF

(de + ef ) - df = 2e || 2e/2 = e || ed - e = d || ef - e = f || f + d =  df

de + ef = 6x + 11 || (6x +11) - df = -3x -6 || 2e/2 = -3x/2 - 3  || (-3x/2 - 3) - de = x/2 + 7||  (- 3x/2 - 3) - ef = -x/2 - 1|| d + f = 6


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3 years ago
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Answer:

The given expression  (\frac{5}{6} a^{9}p^{5})  ^{3} = \frac{125}{216}  \times a^{(27) } \times p^{(15)}

Step-by-step explanation:

Here, the given expression is:  (\frac{5}{6} a^{9}p^{5})  ^{3}

Now, starting from the outer most bracket.

As we know :

(abc)^{n}   = (a)^{n} \times (b)^{n}  \times (c)^{n}

and (a^m)^{n} = a^{(m \times n)}

⇒ (\frac{5}{6} a^{9}p^{5})  ^{3} = (\frac{5}{6})^{3} \times (a^{9})^{3}   \times (p^{5}) ^{3}\\

=\frac{125}{216}  \times (a)^{(9\times3) } \times (p)^{(5 \times 3)}\\= \frac{125}{216}  \times a^{(27) } \times p^{(15)}

Hence, the given expression  (\frac{5}{6} a^{9}p^{5})  ^{3} = \frac{125}{216}  \times a^{(27) } \times p^{(15)}

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