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nexus9112 [7]
3 years ago
5

Rick is on a bicycle trip. Every 444 days he bikes 230 \text{ km}230 km230, space, k, m.If Rick keeps this same pace for 161616

days, how many kilometers will he bike?
Mathematics
1 answer:
Katarina [22]3 years ago
4 0
For this problem, we use the approach of ratio and proportion. Assuming that the given ratio of 444 days per 230 km is constant all throughout, we can determine the number of days or distance as long as one of the two is given. In this case, the solution is as follows:

444 days/230 km = 161616 days/distance
Distance = 83,720 km
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kkurt [141]

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3 years ago
Find the arc-length parametrization of the curve that is the intersection of the elliptic cylinder x 2 + y 2/2 = 1 and the plane
storchak [24]

Answer:

f(\theta) = (cos(\frac{\theta}{\sqrt2}), \sqrt2 sin(\frac{\theta}{\sqrt2}), cos(\frac{\theta}{\sqrt2})-2)

0 ≤ Ф ≤ 4π.

Step-by-step explanation:

since x²+y²/2 = 1, then x²+s² = 1, with s = (y/√2)². Hence, (x,s) = (cos(Ф),sin(Ф)) and (x,y,z) = (cos(Ф),√2 sin(Ф), cos(Ф)-2). This expression evaluated in zero gives as result (1,0,-1). The derivate of this function is (-sin(Ф),√2 cos(Ф), -sen(Ф))

the norm of the derivate is √(sin²(Ф) + 2cos²(Ф)+sin²(Ф)) = √2. In order to make the norm equal to 1, i will divide Ф by √2, so that a √2 is dividing each term after derivating.

We take

f(\theta) = (cos(\frac{\theta}{\sqrt2}), \sqrt2 sin(\frac{\theta}{\sqrt2}), cos(\frac{\theta}{\sqrt2})-2)

Note that

  • f(0) = (1,0,-1)
  • f'(\theta) = (\frac{sin(\frac{\theta}{\sqrt2})}{\sqrt2}, cos(\frac{\theta}{\sqrt2})}, \frac{sin(\frac{\theta}{\sqrt2})}{\sqrt2})

Whose square norm is 1/2cos²(Ф/2)+sen²(Ф/2)+1/2cos²(Ф/2) = 1. This is te parametrization that we wanted.

The values from Ф range between 0 an 4π, because the argument of the sin and cos is Ф/2, not Ф, Ф/2 should range between 0 and 2π.

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3 years ago
The value of 8 in 56,982 is ___ times as large as the value of 8 in 156,408
Rudik [331]

The value of 8 in 56982 is in the tens place value. The value of 8 in 156,408 is in the ones value

Divide

80/8 = 10

The value of 8 in 56,982 is 10 times as large as the value in 8 in 156,408

hope this helps

3 0
3 years ago
Read 2 more answers
Find the inverse laplace<br><br> F(s)=11s/ s^2-12s+52
lions [1.4K]

Answer:

11e^{6t}\cos 4t+\frac{33}{2}e^{6t}\sin 4t

Step-by-step explanation:

We can write \frac{11s}{s^2-12s+52} as follows:

\frac{11s}{s^2-12s+52}\\=11\left [ \frac{s}{s^2-12s+52} \right ]\\=11\left [ \frac{s}{(s-6)^2+16} \right ]\\=11\left [ \frac{s-6+6}{(s-6)^2+16} \right ]\\=11\left [ \frac{s-6}{(s-6)^2+16} \right ]+\frac{66}{(s-6)^2+16}

To find:

L^{-1}\left [ \frac{11s}{s^2-12s+52 \right ]}\\=L^{-1}\left [ 11\left [ \frac{s-6}{(s-6)^2+16} \right ]+\frac{66}{(s-6)^2+16} \right ]

We will use formulae:

L^{-1}\left \{ \frac{s-a}{(s-a)^2+b^2} \right \}=e^{at}\cos bt\\L^{-1}\left \{ \frac{b}{(s-a)^2+b^2} \right \}=e^{at}\sin bt

we get solution as :

L^{-1}\left [ 11\left [ \frac{s-6}{(s-6)^2+16} \right ]+\frac{66}{(s-6)^2+16} \right ]\\=L^{-1}\left [ 11\left [ \frac{s-6}{(s-6)^2+4^2} \right ]+\frac{66}{4}\left [ \frac{4}{(s-6)^2+4^2} \right ] \right ]\\=11e^{6t}\cos 4t+\frac{33}{2}e^{6t}\sin 4t

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