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3 years ago

13

At Garcia's Bike Rentals, it costs 31 to rent a bike for 7 hours. How many hours of bike use does a customer get per dollar?

1 answer:

Vesna [10]3 years ago

5 0

$\tt \dfrac{7}{31}=0.226~hours/dollar$

You might be interested in

You row upstream at a speed of 2 mi/h. You travel the same distance

Reika [66]

Answer:

3.5 m/hr

Step-by-step explanation:

2 + c = 5 -c c = speed of current

2 c = 3 c = current speed = 1.5 m/hr

your speed = 5 - 1.5 (or 2 + 1.5) = 3.5 m/hr

4 0

2 years ago

2/3 * 10/24*-3 1/8* -4/5

nadya68 [22]

Answer: 25/36

Step-by-step explanation:

Multiply:

2/3 * 10/24 = 20/72 = 5/18

-3 1/8 * -4/5 = 5/2

5/18 * 5/2 = 25/36

7 0

3 years ago

1) Which pair of angles are congruent?

gavmur [86]

Answer:The eight angles will together form four pairs of corresponding angles. Angles 1 and 5 constitutes one of the pairs. Corresponding angles are congruent. All angles that have the same position with regards to the parallel lines and the transversal are corresponding pairs e.g. 3 + 7, 4 + 8 and 2 + 6.

Step-by-step explanation:

4 0

3 years ago

Consider the initial value problem y′+5y=⎧⎩⎨⎪⎪0110 if 0≤t<3 if 3≤t<5 if 5≤t<[infinity],y(0)=4. y′+5y={0 if 0≤t<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

$Y(s)=e^{-cs}F(s)\implies y(t)=u(t-c)f(t-c)$

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

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

3 0

4 years ago

A horse gallops 200ft, turns and trots 350ft, turns again and travels 410ft to return to the point he started from.

Andrej [43]

Answer:

I think it's 28,700,000 feet

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers