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schepotkina [342]
3 years ago
11

A gardener installed 42.6 meters of fencing in a week. He installed 13.45 meters on Monday and 9.5 meters on Tuesday . He instal

led the rest of the fence in equal lengths on Wednesday through Friday. How many meters of fencing did he install on each of the last three days?
Mathematics
1 answer:
kvasek [131]3 years ago
8 0
6.55 meters of fencing. Simply subtract 13.45 meters and 9.5 meters from 42.6 meters, and then divide by 3.

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Point C has a _____ abscissa and a _____ ordinate.
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Abcissa is x, and ordinate is y.

In this case the x is positive, and the y is negative.

Therefore, the answer is "positive, negative"

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2 years ago
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LUCKY_DIMON [66]

Answer:

Y = 3x^x is a graph that has exponential growth while y = 3^-x has exponential decay.

Y = 3x^x (-∞, 0) and (∞, ∞).

Y = 3x^-x (-∞, ∞) and (∞, 0).

Step-by-step explanation:

The infinity symbols were being used to represent the x and y values of each graph. I will call y = 3^x "graph 1" and y = 3^-x "graph 2".

When graph 1 had positive ∞ for its x value, its y value was reaching towards positive ∞. When its x was reaching for negative ∞, its y was going for 0.

For graph 2, however, when its x was reaching for positive ∞, its x was reaching for 0. When its x was reaching for negative ∞, its y was going for positive ∞.

Here's an image of the graphs:

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2 years ago
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Vanyuwa [196]

Answer:

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3 years ago
Use the Taylor series you just found for sinc(x) to find the Taylor series for f(x) = (integral from 0 to x) of sinc(t)dt based
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In this question (brainly.com/question/12792658) I derived the Taylor series for \mathrm{sinc}\,x about x=0:

\mathrm{sinc}\,x=\displaystyle\sum_{n=0}^\infty\frac{(-1)^nx^{2n}}{(2n+1)!}

Then the Taylor series for

f(x)=\displaystyle\int_0^x\mathrm{sinc}\,t\,\mathrm dt

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f(x)=\displaystyle\int\sum_{n=0}^\infty\frac{(-1)^nx^{2n}}{(2n+1)!}\,\mathrm dx=C+\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{(2n+1)^2(2n)!}

We have f(0)=0, so C=0 and so

f(x)=\displaystyle\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{(2n+1)^2(2n)!}

which converges by the ratio test if the following limit is less than 1:

\displaystyle\lim_{n\to\infty}\left|\frac{\frac{(-1)^{n+1}x^{2n+3}}{(2n+3)^2(2n+2)!}}{\frac{(-1)^nx^{2n+1}}{(2n+1)^2(2n)!}}\right|=|x^2|\lim_{n\to\infty}\frac{(2n+1)^2(2n)!}{(2n+3)^2(2n+2)!}

Like in the linked problem, the limit is 0 so the series for f(x) converges everywhere.

7 0
3 years ago
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posledela
33 * 3 + 3/3 = 100

I do not believe u have to use all of the operations because it says " at most once ". 
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