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3 years ago

Help how do and is help?

Mathematics

1 answer:

Dimas [21]3 years ago

6 0

Answer:

x = 12

Step-by-step explanation:

The total of all the angle is 180 degrees so just divide 180 by the total number of x's which is 5+4+3+2+1 = 15

180/15 = 12

If you wanted to know how much each angle is just plug in 12 for x.

So first angle is just X so it is 12 degrees.

Second angle is 2X so 2x12 = 24 degrees.

and so on for all the angles.

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George buys some first class and second class stamps to put on some letters. Each first class stamp costs 67p, and each second c

Dafna1 [17]

Answer:

£191.36

Step-by-step explanation:

Sum the parts of the ratio, 1 + 4 = 5 parts

Divide 320 by 5 to find the value of one part of the ratio.

320 ÷ 5 = 64 ← value of 1 part of the ratio , thus

4 parts = 4 × 64 = 256

Thus he bought 64 first class and 256 second class.

Cost = (64 × £0.67) + (256 × £0.58)

= £42.88 + 148.48

= £191.36

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3 years ago

4 questions 50 points

MAXImum [283]

I hope this helps you

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3 years ago

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dlinn [17]

It would be reproduction

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3 years ago

#26 need help please

erma4kov [3.2K]

They will have to buy all three items again in 9 years

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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:

<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}}$