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2 years ago

Can someone please help and explain

Mathematics

1 answer:

dangina [55]2 years ago

7 0

Answer:

$c = 300000 {(1.05)}^{50}$

Step-by-step explanation:

The future value formula is FV=PV(1+i)^n, where the present value PV increases for each period into the future by a factor of 1 + i. value PV increases for each period into the future by a factor of 1 + i.

$fv = pv {(1 + r)}^{n}$

fv = future value

pv = present value

r = annual interest rate ( decimal number)

n= period

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A B means:

SVETLANKA909090 [29]

Answer:

A is a subset of B i believe that's the answer

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2 years ago

Which mathematical sentence most Accurately expresses the information in the problem below?

d1i1m1o1n [39]

Answer:

Don't quote me on this but it's probably C. 12c ≥ 88

Step-by-step explanation:

This is because he packages 88 eggs into cartons of 12.

88/12 is 7.33333... so it makes sense to have a greater or equal amount of eggs. If you multiply 7.333333... by 12, each additional 3 gets you closer to 88 so again, makes sense to have more.

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2 years ago

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

Help me pleasee :) :)

andreyandreev [35.5K]

I know that one is obtuse and acute

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3 years ago

The product of 3/4 and 7/8 is less than 7/8?

luda_lava [24]

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3 years ago