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I am Lyosha [343]

2 years ago

13

Kitty buys hot chocolate sachets. There are 14 hot chocolate sachets in a small box. A small box costs £3.49. Kitty uses 3 hot c

hocolate sachets each day. Work out the how much Kitty spends on hot chocolate sachets in a four-week period.

2 answers:

gogolik [260]2 years ago

5 0

Answer:

24.43

Step-by-step explanation:

first find the price of One sachets

next Find the no. of sachets consumed for four weeks..

and at last the product of the price of one sachet and no. of sachets consumed will give the answer...

Mathematical operation are above...

Svetllana [295]2 years ago

3 0

She spends 73.29 in 4 weeks hope it helps

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WITCHER [35]

Angle O=64

Side m=7.3159882(rounded if needed)

Side n=16.68902(round if needed)

We are given

Angle M which equal 26.

Angle N equals 90 degrees

Side MN which equal 15.

We are asked to find all the missing elements of the triangle.

Angle O can be found using triangle interior theorem so we can add 90+26+x =180 to find the value that satisfies the x value.

$90 + 26 + x = 180$

$116 + x = 180$

$x = 64$

Angle O=64 degrees

We can find side m by doing law of sines.

$\frac{on}{ \sin(26) } = \frac{mn}{ \sin(64) }$

$\frac{on}{ \sin(26) } = \frac{15}{ \sin(64) }$

$on = \frac{15}{ \sin(64) } \times \sin(26)$

which is 7.315(round if neeeded).

We can find side n using pythagorean theorem,

${15}^{2} + {7.3}^{2} = {n}^{2}$

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2 years ago

Jean's bedroom is 14 feet by 13 feet. She has chosen a carpet which costs $28.95 per square yard. This includes installation.

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3 years ago

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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:

<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}}$

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

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

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

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:

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

<u>Taylor Series summation notation</u>:

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

Therefore:

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

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