PLEASE HELP ME

Mathematics

2 answers:

Wewaii [24]3 years ago

5 0

7-5.5=1.5
so each drop leaf is 1.5
2×1.5=3
so 7 when both drop leafs are up
7-3=4
There's your answer

Leno4ka [110]3 years ago

4 0

The length of the table when both leaves are folded up is 7 ft.

The length of the table when one of the leaves are folded up is 5.5 ft.

So if you do:

7 - 5.5 = 1.5

1.5 ft is the length of one of the leaves (remember the two drop leaves are the same size)

How long is the table when no leaves are folded up?

One leaf is 1.5 ft.

Since there are two leaves, the total length of the leaves is 3 ft.

7 - 3 = 4

[7 is the length of the table when both of the leaves are folded up, 3 is the length of both of the leaves]

4 ft is the length of the table when no leaves are folded up

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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:

<u>Taylor series</u> 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}}$