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4 years ago

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Find a formula for the nth partial sum Sn of the following series. If the series is finite, determine the sum. If the series is

infinite, determine if it converges or diverges, and if it converges, determine the sum. Please simplify your solution.

1 answer:

Xelga [282]4 years ago

5 0

Answer:

Step-by-step explanation:

put i=1,2,...,n

$\frac{1}{5}-\frac{1}{6}\\\frac{1}{6}-\frac{1}{7}\\\frac{1}{7}-\frac{1}{8}\\..............\\................\\................\\\frac{1}{n-1}-\frac{1}{n}adding ~vertically~downwards\\S_{n}=\frac{1}{5}-\frac{1}{n}$

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Graph f(x) = x2 + 2x - 3, label the function’s x-intercepts, y-intercept and vertex with their coordinates. Also draw in and lab

Semenov [28]

<h2>Answer with explanation:</h2>

The given function : $f(x)=x^2+2x-3$

Using completing the squares, we have

$f(x)=x^2+2x+1-1-3$ [∵ $(x+1)^2=x^2+2x+1$ ]

$f(x)=(x+1)^2-4$ (1)

Comparing (1) to the standard vertex form $f(x)=(x-h)^2+k$ , the vertex of function is at (h,k)=(-1,-4)

For x-intercept, put f(x)=0 in (1), we get

$0=(x+1)^2-4\\\\\Rightarrow\ (x+1)^2=4$

Square root on both sides, we get

$x+1=\pm2\\\\x+1=-2\ or\ \ x+1=2\\\\=x=-3\ \ or\ x=1$

∴ x intercepts : x= (-3,0) and (1,0)

For y-intercept put x=0 in (1), we get

$f(1)=(1)^2-4=1-4=-3$

∴ y intercept : (0,-3)

Axis of symmetry : $\dfrac{-b}{2a}$

In $f(x)=x^2+2x-3$ , a=1 and b=2

Axis of symmetry= $\dfrac{-2}{2(1)}=-1$

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Step-by-step explanation:

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Step-by-step explanation:

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Please help? Graph <img src="https://tex.z-dn.net/?f=-3x%5E2%2B12y%5E2%3D84" id="TexFormula1" title="-3x^2+12y^2=84" alt="-3x^2+

mariarad [96]

$-3x^2+12y^2=84$
$12y^2=3x^2+84$
$y^2=\dfrac{x^2}4+7$
$y=\pm\sqrt{\dfrac{x^2}4+7}$

For either square root to exist, you require that $\dfrac{x^2}4+7\ge0$ . This is true for all $x$ , since $\dfrac{x^2}4$ is always non-negative. This means the domain of $y$ as a function of $x$ is all real numbers, or $x\in\mathbb R$ or $(-\infty,infty)$ .

Now, because $\dfrac{x^2}4+7$ is non-negative, and the smallest value it can take on is 7, it follows that the minimum value for the positive square root must be $\sqrt7$ , while the maximum value of the negative root must be $-\sqrt7$ . This means the range is $y\in\mathbb R\setminus(-\sqrt7,\sqrt7)$ , or $|y|\ge\sqrt7$ , or $(-\infty,-\sqrt7]\cup[\sqrt7,\infty)$ .

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