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3 years ago

Is the equation y = x3 even, odd, or neither?

Mathematics

1 answer:

ohaa [14]3 years ago

5 0

Let be y =f(x)= x3, f(-x) = (-x)^3 = - x^3 = - f(-x), f(-x) = (-x)^3, f(-x) = - f(-x)
the equation is odd

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If p 0 and p+q=r which statement is true

zhuklara [117]

Answer:p=qbecause they are the same

Step-by-step explanation:

3 0

3 years ago

Lim (n/3n-1)^(n-1) n → ∞

n200080 [17]

Looks like the given limit is

$\displaystyle \lim_{n\to\infty} \left(\frac n{3n-1}\right)^{n-1}$

With some simple algebra, we can rewrite

$\dfrac n{3n-1} = \dfrac13 \cdot \dfrac n{n-9} = \dfrac13 \cdot \dfrac{(n-9)+9}{n-9} = \dfrac13 \cdot \left(1 + \dfrac9{n-9}\right)$

then distribute the limit over the product,

$\displaystyle \lim_{n\to\infty} \left(\frac n{3n-1}\right)^{n-1} = \lim_{n\to\infty}\left(\dfrac13\right)^{n-1} \cdot \lim_{n\to\infty}\left(1+\dfrac9{n-9}\right)^{n-1}$

The first limit is 0, since 1/3ⁿ is a positive, decreasing sequence. But before claiming the overall limit is also 0, we need to show that the second limit is also finite.

For the second limit, recall the definition of the constant, e :

$\displaystyle e = \lim_{n\to\infty} \left(1+\frac1n\right)^n$

To make our limit resemble this one more closely, make a substitution; replace 9/(n - 9) with 1/m, so that

$\dfrac{9}{n-9} = \dfrac1m \implies 9m = n-9 \implies 9m+8 = n-1$

From the relation 9m = n - 9, we see that m also approaches infinity as n approaches infinity. So, the second limit is rewritten as

$\displaystyle\lim_{n\to\infty}\left(1+\dfrac9{n-9}\right)^{n-1} = \lim_{m\to\infty}\left(1+\dfrac1m\right)^{9m+8}$

Now we apply some more properties of multiplication and limits:

$\displaystyle \lim_{m\to\infty}\left(1+\dfrac1m\right)^{9m+8} = \lim_{m\to\infty}\left(1+\dfrac1m\right)^{9m} \cdot \lim_{m\to\infty}\left(1+\dfrac1m\right)^8 \\\\ = \lim_{m\to\infty}\left(\left(1+\dfrac1m\right)^m\right)^9 \cdot \left(\lim_{m\to\infty}\left(1+\dfrac1m\right)\right)^8 \\\\ = \left(\lim_{m\to\infty}\left(1+\dfrac1m\right)^m\right)^9 \cdot \left(\lim_{m\to\infty}\left(1+\dfrac1m\right)\right)^8 \\\\ = e^9 \cdot 1^8 = e^9$

So, the overall limit is indeed 0:

$\displaystyle \lim_{n\to\infty} \left(\frac n{3n-1}\right)^{n-1} = \underbrace{\lim_{n\to\infty}\left(\dfrac13\right)^{n-1}}_0 \cdot \underbrace{\lim_{n\to\infty}\left(1+\dfrac9{n-9}\right)^{n-1}}_{e^9} = \boxed{0}$

7 0

3 years ago

Which is a parallelogram with four congruent sides whose diagonals bisect each other and intersect at four right angles?

Charra [1.4K]

Answer:

I believe that the answer is B. Rhombus.

Step-by-step explanation:

A rhombus has the properties of a parallelogram and the diagonals intersect at right angles.

Hope this helps!

3 0

3 years ago

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Boyle’s law states that the volume of a gas varies inversely with applied pressure. Suppose the pressure on 60 cubic meters of g

tatyana61 [14]

$\bf \qquad \qquad \textit{inverse proportional variation}
\\\\
\textit{\underline{y} varies inversely with \underline{x}}\qquad \qquad y=\cfrac{k}{x}\impliedby 
\begin{array}{llll}
k=constant\ of\\
\qquad variation
\end{array}\\\\
-------------------------------$

$\bf \stackrel{\textit{the volume of a gas varies inversely with applied pressure}}{v=\cfrac{k}{p}}
\\\\\\
\textit{we also know that }
\begin{cases}
v=60\\
p=1
\end{cases}\implies 60=\cfrac{k}{1}\implies 60=k
\\\\\\
\boxed{v=\cfrac{60}{p}}
\\\\\\
\stackrel{\textit{now if we apply 3 atmospheres, p=3, what is \underline{v}?}}{v=\cfrac{60}{3}}$

8 0

3 years ago

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Find the equation of the line passing through the points 1,-5) and (9,11). y = 2x + [?]

Naily [24]

Answer:

y = 2x - 7

Step-by-step explanation:

Looks like we already have the slope of this line: It is 2. Working with the point (1, -5), we have x = 1 and y = -5 and can from this info easily find the y-intercept, b:

y = mx + b becomes

y = 2x + b, which in turn becomes

-5 = 2(1) + b, or

b = -7,

and so the desired equation is y = 2x - 7

3 0

3 years ago