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3 years ago

5

A circular running track is ¼ mile long. Elena runs on this track, completing each lap in 1/16 of an hour. What is Elena’s runni

ng speed? Include the unit of measure?

1 answer:

AVprozaik [17]3 years ago

3 0

Given that,

A circular running track is ¼ mile long.

Elena runs on this track, completing each lap in 1/16 of an hour

To find,

Elena’s running speed.

Solution,

Formula used :

Speed = distance/time

$v=\dfrac{\dfrac{1}{4}}{\dfrac{1}{16}}\\\\=\dfrac{1}{4}\times 16\\\\=4\ mph$

So, Elena’s running speed is 4 mph.

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Answer:

$y(x)=e^{\frac{-x^{2}}{2}+2 }+1$

Step-by-step explanation:

We have that:

$y(x)=2+\int\limits^x_2 {[t-ty(t)]} \, dt$ (Equation 1)

To resolve this integral equation, we need to use the second Fundamental Theorem of Calculus, which says:

$\frac{d}{dx} [\int\limits^x_a {f(t)} \, dt]=f(x)$

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$\frac{dy}{dx} =\frac{d}{dx} [2+\int\limits^x_2 {[t-ty(t)]} \, dt]$

$\frac{dy}{dx} =\frac{d(2)}{dx}+\frac{d}{dx} [\int\limits^x_2 {[t-ty(t)]} \, dt]$

We know that the derivate for a constant value is zero. And,

$\frac{dy}{dx}=\frac{d}{dx} [\int\limits^x_2 {t} \, dt]-\frac{d}{dx}[\int\limits^x_2 {ty(t)} \, dt]$ (Equation 2)

Using the second Fundamental Theorem of Calculus we know that:

$\frac{d}{dx} [\int\limits^x_2 {t} \, dt]=x\\ \frac{d}{dx} [\int\limits^x_2 {ty(t)} \, dt]=xy(x)$

So, we need to replace those equations in equation 2, and we obtain:

$\frac{dy}{dx} =x-xy(x)$ (Equation 3)

Now, we are going to resolve the equation 3 as a normal equation. So, we need to joint the same variables. I mean, variable y on one side and variable x on other side, as follows:

$\frac{dy}{dx} =x(1-y)\\\frac{dy}{(1-y)} =xdx\\\frac{dy}{y-1} =-xdx$

And, we integrate each side of the equation to obtain:

$ln|y-1|=-\frac{x^{2} }{2} +C$ (Equation 4)

Now, we need to find the value of the constant C. And we know that we can find one point of the equation, replacing x=2 in equation 1, because the integral becomes zero, so:

$y(2)=2+\int\limits^2_2 {t-ty(t)} \, dt=2$

And, we replace the value of y when x=2 in equation 4 and we obtain,

$ln|2-1|=-\frac{2^{2} }{2} +C\\C=2$

So, Equation 4 is:

$ln|y-1|=-\frac{x^{2} }{2} +2$ (Equation 4')

Now, we need to clear the y variable from Equation 4', (we are going to asumme that y-1>0),

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8 0

3 years ago

Crank

Answer:

C = $\frac{963}{92}$

Step-by-step explanation:

given A varies directly as B and inversely as C then the equation relating them is

A = $\frac{kB}{C}$ ← k is the constant of variation

to find k use the condition A = 6 , B = 10 , C = 15 , then

6 = $\frac{x10k}{15}$ ( multiply both sides by 15 )

90 = 10k ( divide both sides by 10 )

9 = k

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when A = 92 and B = 107 , then

92 = $\frac{9(107)}{C}$ ( multiply both sides by C )

92C = 963 ( divide both sides by 92 )

C = $\frac{963}{92}$

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2 years ago

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deff fn [24]

Answer:

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Step-by-step explanation:

Lets take (0,-6) and (1,-10)

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= ((-10+6)/(1-0)

= -4/1

so, m = -4

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or, -10=-4×1+c

or, -10 = -4+c

so, c = -6

now

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or, y = -4x-6

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3 years ago

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