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3 years ago

Which is the estimate of 2584?

1 answer:

8 0

Not quite sure about the answer...maybe we should look for some better like problems....overall, Good luck. I'm sorry I tried to think the best I can..

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Use the definition of a Taylor series to find the first three non zero terms of the Taylor series for the given function centere

Ket [755]

Answer:

$e^{4x}=e^4+4e^4(x-1)+8e^4(x-1)^2+...$

$\displaystyle e^{4x}=\sum^{\infty}_{n=0} \dfrac{4^ne^4}{n!}(x-1)^n$

Step-by-step explanation:

Taylor series expansions of f(x) at the point x = a

$\text{f}(x)=\text{f}(a)+\text{f}\:'(a)(x-a)+\dfrac{\text{f}\:''(a)}{2!}(x-a)^2+\dfrac{\text{f}\:'''(a)}{3!}(x-a)^3+...+\dfrac{\text{f}\:^{(r)}(a)}{r!}(x-a)^r+...$

This expansion is valid only if $\text{f}\:^{(n)}(a)$ exists and is finite for all $n \in \mathbb{N}$ , and for values of x for which the infinite series converges.

$\textsf{Let }\text{f}(x)=e^{4x} \textsf{ and }a=1$

$\text{f}(x)=\text{f}(1)+\text{f}\:'(1)(x-1)+\dfrac{\text{f}\:''(1)}{2!}(x-1)^2+...$

$\boxed{\begin{minipage}{5.5 cm}\underline{Differentiating $e^{f(x)}$}\\\\If $y=e^{f(x)}$, then $\dfrac{\text{d}y}{\text{d}x}=f\:'(x)e^{f(x)}$\\\end{minipage}}$

$\text{f}(x)=e^{4x} \implies \text{f}(1)=e^4$

$\text{f}\:'(x)=4e^{4x} \implies \text{f}\:'(1)=4e^4$

$\text{f}\:''(x)=16e^{4x} \implies \text{f}\:''(1)=16e^4$

Substituting the values in the series expansion gives:

$e^{4x}=e^4+4e^4(x-1)+\dfrac{16e^4}{2}(x-1)^2+...$

Factoring out e⁴:

$e^{4x}=e^4\left[1+4(x-1)+8}(x-1)^2+...\right]$

Taylor Series summation notation:

$\displaystyle \text{f}(x)=\sum^{\infty}_{n=0} \dfrac{\text{f}\:^{(n)}(a)}{n!}(x-a)^n$

Therefore:

$\displaystyle e^{4x}=\sum^{\infty}_{n=0} \dfrac{4^ne^4}{n!}(x-1)^n$

7 0

1 year ago

Solve using the elimination method: X-3y=7 3x + 3y=9

sergij07 [2.7K]

Answer:

$x=4\\\\y=-1$

Step-by-step explanation:

Given Equation:

$x-3y=7$ Equation:1

$3x + 3y=9$ Equation:2

Dividing Equation:2 by '3' both the sides:

$x+y=3$

or $x=3-y$ Equation:3

Putting the vale of 'x' in Equation:1

$x-3y=7$

$(3-y)-3y=7\\\\3-y-3y=7\\\\3-4y=7\\\\$

Subtracting '3' both sides

$-4y=7-3\\\\-4y=4$

$y=\frac{4}{-4}$

$y=-1$

Putting value of 'y' in Equation:3

$x=3-y$

$x=3-(-1)\\\\x=3+1\\\\x=4$

The solution of the equations is :

$x=4\\\\y=-1$

7 0

3 years ago

The equations x minus 2 y = 1, 3 x minus y = negative 1, x + 2 y = negative 1, and 3 x + y = 1 are shown on the graph below. On

Basile [38]

Answer: (a) x + 2y = -1 and 3x + y = 1

Step-by-step explanation:

I am not sure what the purpose was for the colored lines but I included them on the graph (below).

6 0

3 years ago

Read 2 more answers

Chloe and Libby want to tie Ryan to a bunch of approximately spherical helium balloons of diameter 0.3m. The volume of each ball

dem82 [27]

Answer: Aproximately 2,525 balloons

Step-by-step explanation:

1. Find the volume of a balloon with the formula given in the problem, where $r$ is the radius ( $r=\frac{0.3m}{2}=0.15m$ ), then:

$V=4(0.15m)^{3}=0.0135m^{3}$

2. Convert the volume from m³ to lliters by multiplying it by 1,000:

$V=0.0135m^{3}*1000=13.5L$

3. You know that that 1 liter of helium can lift 1 gram and that Ryan weighs 5 stone and 5 pounds. So you must make the following conversions:

1 g=0.0022 lb

From 5 stones to pounds

$(5stone)(\frac{14pounds}{1stone})=70pounds=70lbs$

4. Then Ryan's weigh is:

5lb+70lb=75lb

5. Then, if 1 liter of helium can lift 0.0022 lb, to lift 75 lb (which is the weight of Ryan) they need:

$\frac{75lb*1L}{0.0022lb}=34,090.90L$

6. Then, to calculate the aproximated number of balloons they need to make him float (which can call $n$ ), you must divide the liters of helium needed to lift the weight of Ryan by the volume of a balloon, then the result is: