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Leokris [45]
3 years ago
15

Please help me ASAP easy problem giving brainlist!!

Mathematics
1 answer:
EleoNora [17]3 years ago
3 0
It’s the very first one
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What is the inverse of the function below?<br> f(x) = 3^x
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log ^3 (x)

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See the image below:)

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3 years ago
Help please!!!!!!!!!!!!!
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√20,

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1. Find the midpoint of the segment with endpoints of 6 3i and −12 − 5i. I.
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3 years ago
Lindsey is considering purchasing an item that costs $860 and has a discount of 15 % . What is the amount of discount on
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$129

Step-by-step explanation:

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4 0
4 years ago
For how many real values of x is <img src="https://tex.z-dn.net/?f=%20%5Csqrt%7B120-%5Csqrt%7Bx%7D%7D%20" id="TexFormula1" title
leva [86]

A good place to start is to set \sqrt{x} to y. That would mean we are looking for \sqrt{120-y} to be an integer. Clearly, y\leq 120, because if y were greater the part under the radical would be a negative, making the radical an imaginary number, not an integer. Also note that since \sqrt{x} is a radical, it only outputs values from [0,\infty], which means y is on the closed interval: [0,120].

With that, we don't really have to consider y anymore, since we know the interval that \sqrt{x} is on.

Now, we don't even have to find the x values. Note that only 11 perfect squares lie on the interval [0,120], which means there are at most 11 numbers that x can be which make the radical an integer. All of the perfect squares are easily constructed. We can say that if k is an arbitrary integer between 0 and 11 then:

\sqrt{120-\sqrt{x}}=k \implies \\ \sqrt{x}=k^2-120 \implies\\ x=(k^2-120)^2

Which is strictly positive so we know for sure that all 11 numbers on the closed interval will yield a valid x that makes the radical an integer.

5 0
3 years ago
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