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3 years ago

Lorenzo rides his bike at a rate of 5 yards per second. About how many miles per hour can Lorenzo ride his bike?

Mathematics

1 answer:

nikitadnepr [17]3 years ago

4 0

5280 ft =1 mi
hour =3600seconds 5280/3 =1760 yd= 1 mi
5yd/sec=x/3600
x=18000 yd/1760yd
x=10.227 miles

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once a week you babysit your neighbor’s toddler after school, usually going to a local playground. you notice that each swing on

Natalka [10]

You didn't include the formula.

Given that there is no data about the mass, I will suppose that the formula is that of the simple pendulum (which is only valid if the mass is negligible).

Any way my idea is to teach you how to use the formula and you can apply the procedure to the real formula that the problem incorporates.

Simple pendulum formula:

Period = 2π √(L/g)

Square both sides

Period^2 = (2π)^2 L/g

L = [Period / 2π)^2 * g

Period = 3.1 s
2π ≈ 6.28
g ≈ 10 m/s^2

L = [3.1s/6.28]^2 * 10m/s^2 =2.43 m

I hope this helps you.

6 0

3 years ago

Based on what you know what is the measure of m

Bond [772]

Answer:

59 degrees, 44 plus 77 is 121, subtract from 180

4 0

3 years ago

Stephen, Nick and Brian share £90 in a ratio 1:2:2. How much money does each person get?

malfutka [58]

Answer:

Stephen: £36

Nick: £36

Brian: £18

Step-by-step explanation:

2+2+1= 5

90/5= 18

18x2= 36

18x2=36

18x1=18

Hope it helps

4 0

3 years ago

Please help will give brainliest it’s for a test<br><br> Round 29.20 to 30 <br> Show work please

djverab [1.8K]

Answer:

you cant round that to 30

Step-by-step explanation

29.20

29.2 2 is less than 5

so you cant round you can only round if the .tens place is 5 or more

4 0

3 years ago

I'm having trouble with #2. I've got it down to the part where it would be the integral of 5cos^3(pheta)/sin(pheta). I'm not sur

Butoxors [25]

$\displaystyle\int\frac{\sqrt{25-x^2}}x\,\mathrm dx$

Setting $x=5\sin\theta$ , you have $\mathrm dx=5\cos\theta\,\mathrm d\theta$ . Then the integral becomes

$\displaystyle\int\frac{\sqrt{25-(5\sin\theta)^2}}{5\sin\theta}5\cos\theta\,\mathrm d\theta$
$\displaystyle\int\sqrt{25-25\sin^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$
$\displaystyle5\int\sqrt{1-\sin^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$
$\displaystyle5\int\sqrt{\cos^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$

Now, $\sqrt{x^2}=|x|$ in general. But since we want our substitution $x=5\sin\theta$ to be invertible, we are tacitly assuming that we're working over a restricted domain. In particular, this means $\theta=\sin^{-1}\dfrac x5$ , which implies that $\left|\dfrac x5\right|\le1$ , or equivalently that $|\theta|\le\dfrac\pi2$ . Over this domain, $\cos\theta\ge0$ , so $\sqrt{\cos^2\theta}=|\cos\theta|=\cos\theta$ .

Long story short, this allows us to go from

$\displaystyle5\int\sqrt{\cos^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$

to

$\displaystyle5\int\cos\theta\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$
$\displaystyle5\int\dfrac{\cos^2\theta}{\sin\theta}\,\mathrm d\theta$

Computing the remaining integral isn't difficult. Expand the numerator with the Pythagorean identity to get

$\dfrac{\cos^2\theta}{\sin\theta}=\dfrac{1-\sin^2\theta}{\sin\theta}=\csc\theta-\sin\theta$

Then integrate term-by-term to get

$\displaystyle5\left(\int\csc\theta\,\mathrm d\theta-\int\sin\theta\,\mathrm d\theta\right)$
$=-5\ln|\csc\theta+\cot\theta|+\cos\theta+C$

Now undo the substitution to get the antiderivative back in terms of $x$ .

$=-5\ln\left|\csc\left(\sin^{-1}\dfrac x5\right)+\cot\left(\sin^{-1}\dfrac x5\right)\right|+\cos\left(\sin^{-1}\dfrac x5\right)+C$

and using basic trigonometric properties (e.g. Pythagorean theorem) this reduces to

$=-5\ln\left|\dfrac{5+\sqrt{25-x^2}}x\right|+\sqrt{25-x^2}+C$

4 0

3 years ago

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