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coldgirl [10]
2 years ago
8

Lauren and minki are in a bike race. Lauren bikes at 18 miles per hour. Minki bikes at 12 miles per hour and has a 6-mile head s

tart. Write the system of equations that represents each persons distance. Let d represent the distance and t represent the time
Mathematics
1 answer:
Alex787 [66]2 years ago
5 0

Answer:

For Lauren - d = 18t

For Minki - d = 6 + 12t

Step-by-step explanation:

Given The speed of Lauren  is 18 miles per hour and the speed of Minki is 12 miles per hour

Also Given that Minki got a head start of 6 miles

For Lauren

Speed=\frac{distance}{time}

18=\frac{d}{t}

d = 18t

For Minki

Speed=\frac{distance}{time}

12=\frac{d}{t}

d = 12t

Also given that he had a 6 miles headstart

Therefore the equation becomes

d = 6 + 12t

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3/4

Step-by-step explanation:

2/3a-1/6=1/3

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Suppose that F(x) = x^3 and G(X) = -3x^3. Which statement best compares the
Alik [6]

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D. The graph of G(x) is the graph of F(x) flipped over the y-axis and

stretched vertically.

Step-by-step explanation:

since there is a negative, there must be a reflection. the reflection is over the y-axis because it is outside the x-value (-F(x) not F(-x))

the same goes for the vertical stretch. we know it is a vertical stretch because the 3 is outside the x-value (3F(x) not F(3x))

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Find the Fourier series of f on the given interval. f(x) = 1, ?7 < x < 0 1 + x, 0 ? x < 7
Zolol [24]
f(x)=\begin{cases}1&\text{for }-7

The Fourier series expansion of f(x) is given by

\dfrac{a_0}2+\displaystyle\sum_{n\ge1}a_n\cos\frac{n\pi x}7+\sum_{n\ge1}b_n\sin\frac{n\pi x}7

where we have

a_0=\displaystyle\frac17\int_{-7}^7f(x)\,\mathrm dx
a_0=\displaystyle\frac17\left(\int_{-7}^0\mathrm dx+\int_0^7(1+x)\,\mathrm dx\right)
a_0=\dfrac{7+\frac{63}2}7=\dfrac{11}2

The coefficients of the cosine series are

a_n=\displaystyle\frac17\int_{-7}^7f(x)\cos\dfrac{n\pi x}7\,\mathrm dx
a_n=\displaystyle\frac17\left(\int_{-7}^0\cos\frac{n\pi x}7\,\mathrm dx+\int_0^7(1+x)\cos\frac{n\pi x}7\,\mathrm dx\right)
a_n=\dfrac{9\sin n\pi}{n\pi}+\dfrac{7\cos n\pi-7}{n^2\pi^2}
a_n=\dfrac{7(-1)^n-7}{n^2\pi^2}

When n is even, the numerator vanishes, so we consider odd n, i.e. n=2k-1 for k\in\mathbb N, leaving us with

a_n=a_{2k-1}=\dfrac{7(-1)-7}{(2k-1)^2\pi^2}=-\dfrac{14}{(2k-1)^2\pi^2}

Meanwhile, the coefficients of the sine series are given by

b_n=\displaystyle\frac17\int_{-7}^7f(x)\sin\dfrac{n\pi x}7\,\mathrm dx
b_n=\displaystyle\frac17\left(\int_{-7}^0\sin\dfrac{n\pi x}7\,\mathrm dx+\int_0^7(1+x)\sin\dfrac{n\pi x}7\,\mathrm dx\right)
b_n=-\dfrac{7\cos n\pi}{n\pi}+\dfrac{7\sin n\pi}{n^2\pi^2}
b_n=\dfrac{7(-1)^{n+1}}{n\pi}

So the Fourier series expansion for f(x) is

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2 years ago
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