How do the last four lines of the poem contribute to the poems overall structure. help me pls!

Mathematics

2 answers:

dusya [7]1 year ago

5 0

A warm june day

we decided to do something new.

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The path of a ball kicked from the ground can be modeled by the equation y=−1/5(x−5)(x−20), where x and y are measured in feet.

Helen [10]

Answer:

Step-by-step explanation:

This is already factored for the zeros of the quadratic. If x - 5 = 0 and x - 20 = 0 then x = 5 and 20. The ball was kicked 5 feet out and landed at 20 feet down the field. The difference is 15 feet. That's how far the ball traveled from its initial position on the ground to its landing place.

So 15 feet from where it was kicked.

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6. A tank contains 100-gallon of pure water. At time t = 0, a solution containing 2 lb of salt per gallon flows into the tank at

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Answer:

The answer is below

Step-by-step explanation:

Let Q represent the amount of salt in the tank at time t.

$\frac{dQ}{dt}= flow\ in - flow\ out\\ \\flow\ in=3\ gal/min*2\ lb/gal=6\ lb/min\\\\net\ gain\ in\ tank\ volume=3-4=-1, henceflow\ out= \frac{4Q}{100-t} \\\\\frac{dQ}{dt}= 6-\frac{4Q}{100-t} \\\\\frac{dQ}{dt}+ \frac{4Q}{100-t}=6\\\\The \ integrating\ factor\ is:\\\\IF=e^{\int\limits {\frac{4}{100-t} } \, dt }=e^{-4\int\limits {\frac{-1}{100-t}}=e^{-4ln(100-t)}=(100-t)^{-4}}\\\\Multiplying\ through \ by\ IF: \\\\(100-t)^{-4}\frac{dQ}{dt}+ (100-t)^{-4}\frac{4Q}{100-t}=6(100-t)^{-4}\\\\$

$Integrating:\\\\A(100-t)^{-4}=-2(100-t)^{-3}+c\\\\A=-2(100-t)+\frac{c}{(100-t)^{-4}} \\\\at, t=0,A=0\\\\0=-2(100-0)+\frac{c}{(100-0)^{-4}}\\\\c=0.02\\\\A=-2(100-t)+\frac{0.02}{(100-t)^{-4}}$

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Then multiply 4 $\frac{2}{4}$ and 3 $\frac{1}{4}$
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