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strojnjashka [21]
3 years ago
8

Usain Bolt of Jamaica set a world record in 2009 for the 200- meter dash with a time of 19.19 seconds. What was his average spee

d
Mathematics
1 answer:
kirill [66]3 years ago
6 0

Answer:

His average speed was of 10.42 m/s

Step-by-step explanation:

We use the following relation to solve this question:

v = \frac{d}{t}

In which v is the velocity(in m/s), d is the distance(in meters), and t is the time, in seconds.

Usain Bolt of Jamaica set a world record in 2009 for the 200- meter dash with a time of 19.19 seconds.

This means that d = 200, t = 19.19

So his average speed was of:

v = \frac{d}{t} = \frac{200}{19.19} = 10.42

His average speed was of 10.42 m/s

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(200-50) + 50 * 2 in hurry
Neko [114]

Answer: 400

Step-by-step explanation:

200-50=150+50=200*2=400

Are you just posting your questions for an assignment?

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2 years ago
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Two times the greater of two consecutive integers is 9 less than three times the lesser integer. What are the integers?
lara [203]

Answer:

11 and 12

Step-by-step explanation:

x=smaller

y=larger

2y=3x-9    x+1=y

2(x+1)=3x-9

2x+2=3x-9

11=x

                11+1=y

                    12=y

3 0
3 years ago
I NEED HELP ASAP PLEASE
skad [1K]
A^2+b^2=c^2 so c^2=12^2+16^2 which then simplifies to c^2=144+256 then simplify that to c^2=400. After that take the square root of c^2 and the square root of 400. So your answer is c=20
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4 years ago
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Your answer would be C-64.

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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

7 0
2 years ago
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