Consider the data set.

Mathematics

1 answer:

denis-greek [22]3 years ago

8 0

Answer:

Median: 9

1 3 6 6 9 10 11 12 31

Step-by-step explanation:

To find the median, you have to organize your terms from least to greatest. And then find the center of your terms, which would be 9.

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Derivative of these questions

shtirl [24]

Answer: $\bold{y'=\dfrac{1}{2(a-b)^2}\bigg(\dfrac{1}{\sqrt{x+a}}+\dfrac{1}{\sqrt{x+b}}\bigg)}$

Step-by-step explanation:

$y=\dfrac{1}{\sqrt{x+a}-\sqrt{x+b}}\\\\\\\text{Rationalize the denominator:}\\y=\dfrac{1}{\sqrt{x+a}-\sqrt{x+b}}\bigg(\dfrac{\sqrt{x+a}+\sqrt{x+b}}{\sqrt{x+a}+\sqrt{x+b}}\bigg)\\\\\\y=\dfrac{\sqrt{x+a}+\sqrt{x+b}}{x+a-x-b}\\\\\\y=\dfrac{\sqrt{x+a}+\sqrt{x+b}}{a-b}\\\\\\\text{Apply the derivative (derivative of the numerator (top) divided by the}\\\text{denominator (bottom) squared).}$

$\text{Derivative of }\sqrt{x+a}\quad \rightarrow \quad (x+a)^{\frac{1}{2}}\quad = \quad \dfrac{1}{2}(x + a)^{-\frac{1}{2}}\\\\\text{Derivative of }\sqrt{x+b}\quad \rightarrow \quad (x+b)^{\frac{1}{2}}\quad = \quad \dfrac{1}{2}(x + b)^{-\frac{1}{2}}\\\\\\\text{Derivative of}\ \dfrac{\sqrt{x+a}+\sqrt{x+b}}{a-b}:\\\\= \dfrac{\dfrac{1}{2}(x + a)^{-\frac{1}{2}}+\dfrac{1}{2}(x + b)^{-\frac{1}{2}}}{(a-b)^2}$

$\text{Factor out }\dfrac{1}{2}\ \text{from the numerator:}\\\\\dfrac{(x+a)^{-\frac{1}{2}}+(x+b)^{-\frac{1}{2}}}{2(a-b)^2}\\\\\\\text{Move the terms with negative exponents to the denominator:}\\\\=\dfrac{1}{2(a-b)^2}\bigg(\dfrac{1}{\sqrt{x+a}}+\dfrac{1}{\sqrt{x+b}}\bigg)$

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find a counterexample to show that the following statment is false. if a number is multipled by itself, the result is even. choo

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B. We can rule out choice B since we're not multiplying here

C. We are multiplying the same value with itself but the result is even. We can rule this out.

D. This is a proper counter example. We have the same number (5) multiplied by itself leading to a result that is not even. It contradicts the initial claim.

Answer: Choice D

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