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Mekhanik [1.2K]
1 year ago
8

3) There are 16 cups in a gallon. The equation c=16g gives the number of cups in terms of the number of gallons. Write another e

quation for this situation, giving the number of gallons in terms of the number of cups:
Mathematics
1 answer:
suter [353]1 year ago
3 0

We know the equation

c=16g

gives the number of cups if we know the numbe rog gallons.

If we want an equation that gives the number of gallons if we know the number of cups we just have to solve hte equation for g, that is:

g=\frac{c}{16}

Therefore, the equation we are looking for is:

g=\frac{c}{16}

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gtnhenbr [62]

Answer:

(-12, -12)

Step-by-step explanation:

Since the scale factor is 3, you would multiply the x-coordinate and y-coordinate by it.

-4×3=-12

Hope this helps!!

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3 years ago
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ch4aika [34]

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First answer.

Step-by-step explanation:

We know that a complete rotation would be 360°. If the figure is being rotated 90°, this is a quarter of a complete rotation.

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3 years ago
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Consider the initial value problem y′+5y=⎧⎩⎨⎪⎪0110 if 0≤t&lt;3 if 3≤t&lt;5 if 5≤t&lt;[infinity],y(0)=4. y′+5y={0 if 0≤t&lt;311 i
rosijanka [135]

It looks like the ODE is

y'+5y=\begin{cases}0&\text{for }0\le t

with the initial condition of y(0)=4.

Rewrite the right side in terms of the unit step function,

u(t-c)=\begin{cases}1&\text{for }t\ge c\\0&\text{for }t

In this case, we have

\begin{cases}0&\text{for }0\le t

The Laplace transform of the step function is easy to compute:

\displaystyle\int_0^\infty u(t-c)e^{-st}\,\mathrm dt=\int_c^\infty e^{-st}\,\mathrm dt=\frac{e^{-cs}}s

So, taking the Laplace transform of both sides of the ODE, we get

sY(s)-y(0)+5Y(s)=\dfrac{e^{-3s}-e^{-5s}}s

Solve for Y(s):

(s+5)Y(s)-4=\dfrac{e^{-3s}-e^{-5s}}s\implies Y(s)=\dfrac{e^{-3s}-e^{-5s}}{s(s+5)}+\dfrac4{s+5}

We can split the first term into partial fractions:

\dfrac1{s(s+5)}=\dfrac as+\dfrac b{s+5}\implies1=a(s+5)+bs

If s=0, then 1=5a\implies a=\frac15.

If s=-5, then 1=-5b\implies b=-\frac15.

\implies Y(s)=\dfrac{e^{-3s}-e^{-5s}}5\left(\frac1s-\frac1{s+5}\right)+\dfrac4{s+5}

\implies Y(s)=\dfrac15\left(\dfrac{e^{-3s}}s-\dfrac{e^{-3s}}{s+5}-\dfrac{e^{-5s}}s+\dfrac{e^{-5s}}{s+5}\right)+\dfrac4{s+5}

Take the inverse transform of both sides, recalling that

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where F(s) is the Laplace transform of the function f(t). We have

F(s)=\dfrac1s\implies f(t)=1

F(s)=\dfrac1{s+5}\implies f(t)=e^{-5t}

We then end up with

y(t)=\dfrac{u(t-3)(1-e^{-5t})-u(t-5)(1-e^{-5t})}5+5e^{-5t}

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nlexa [21]
8.60 cm is the answer to your question and i used calculator.net for help
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