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lions [1.4K]
3 years ago
8

At what points does the helix r(t) = sin t, cos t, t intersect the sphere x2 + y2 + z2 = 65? (round your answers to three decima

l places. if an answer does not exist, enter dne.)
Mathematics
1 answer:
Firdavs [7]3 years ago
7 0
\mathbf r(t)=\langle x(t),y(t),z(t)\rangle=\langle\sin t,\cos t,t\rangle

x^2+y^2+z^2=\sin^2t+\cos^2t+t^2=65
\implies t^2=64
\implies t=\pm8
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Jamel has a cell phone that has unlimited minutes and texts for $45 a month, but he has to pay $5 for each gigabyte of data he u
balandron [24]

Answer:

I think it's 4 gigabytes

Step-by-step explanation:

He only wants to pay 65$ and he already spends 45$ on his phone already. So you only have 20 dollars until his grand totaly of 65$ so 5x4= 20 so he can only use 4 gigabytes. Does that make sense? Hope I helped eheh

8 0
3 years ago
Chioe stops for lunch in a town that
mina [271]

Answer:

B, C, E

Step-by-step explanation:

Tips = $20 × 15%

          $20 × \frac{15}{100}

          $3

Total = $20

Without tips = $17

Things that can buy for $17 are B, C, E

6 0
3 years ago
9. Aaron has a lemonade business. If he sells each glass of lemonade for $2.75, and he sold 11
marin [14]

Answer:

$27.75

Step-by-step explanation:

Multiply $2.25 by 11 to get the answer

4 0
3 years ago
Read 2 more answers
a band of 45 ewoks crash-landed in the forest last night. this sounds like a small problem, but the population will grow at the
kogti [31]

Given Information:

Starting population = P₀ = 45

rate of growth = 22%

Required Information:

Population every five years from this year to the year 2050 = ?

Answer:

Year \: 2020 = P(0) = 45e^{0} = 45\\\\Year \: 2025 = P(5) = 45e^{0.22*5} = 135\\\\Year \: 2030 = P(10) = 45e^{0.22*10} = 406\\\\Year \: 2035 = P(15) = 45e^{0.22*15} = 1,220\\\\Year \: 2040 = P(20) = 45e^{0.22*20} = 3,665\\\\Year \: 2045 = P(25) = 45e^{0.22*25} = 11,011\\\\Year \: 2050 = P(30) = 45e^{0.22*30} = 33,079\\\\

Step-by-step explanation:

The population growth can be modeled as an exponential function,

P(t) = P_0e^{rt}

Where P₀ is the starting population, r is the rate of growth of the population and t is the time in years.

We are given that starting population of 45 and growth rate of 22%

P(t) = 45e^{0.22t}

Assuming that the starting year is 2020,

Year \: 2020 = P(0) = 45e^{0} = 45\\\\Year \: 2025 = P(5) = 45e^{0.22*5} = 135\\\\Year \: 2030 = P(10) = 45e^{0.22*10} = 406\\\\Year \: 2035 = P(15) = 45e^{0.22*15} = 1,220\\\\Year \: 2040 = P(20) = 45e^{0.22*20} = 3,665\\\\Year \: 2045 = P(25) = 45e^{0.22*25} = 11,011\\\\Year \: 2050 = P(30) = 45e^{0.22*30} = 33,079\\\\

Therefore, the starting population of ewoks was 45 in 2020 and increased to 33,079 by 2050 in a time span of 30 years.

8 0
3 years ago
3
guapka [62]
Let p = the number of pennies
let q = the number of quarters

p + q = 23
0.01p + 0.25q = 3.11
3 0
2 years ago
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