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Nat2105 [25]
2 years ago
8

Two hot air balloons are flying above a park. One balloon started at a height of 3,000 feet above the ground and is decreasing i

n height at a rate
of 40 feet per minute. The second balloon is rising at a rate of 50 feet per minute after beginning from a height of 1.200 feet above the ground.
Given that his the height of the balloons after m minutes, determine which system of equations represents this situation.

Mathematics
1 answer:
Stels [109]2 years ago
8 0

Answer:

a

Step-by-step explanation:

its a

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Consider 7 x 10 ^ 3 write a pattern to find the value of the expression
Lelechka [254]
7 x 10^1 = 70
7 x 10^2 = 700
7 x 10^3 = 7000
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Goran went for a drive in his new car. he drove for 4 hours at a speed of 25.6 miles per hour. for how many miles did he drive?
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Distance = Speed * Time

D = 25.6 * 4

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Power series of y''+x^2y'-xy=0
Ray Of Light [21]
Assuming we're looking for a power series solution centered around x=0, take

y=\displaystyle\sum_{n\ge0}a_nx^n
y'=\displaystyle\sum_{n\ge1}na_nx^{n-1}
y''=\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}

Substituting into the ODE yields

\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}+\sum_{n\ge1}na_nx^{n+1}-\sum_{n\ge0}a_nx^{n+1}=0

The first series starts with a constant term; the second series starts at x^2; the last starts at x^1. So, extract the first two terms from the first series, and the first term from the last series so that each new series starts with a x^2 term. We have

\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}=2a_2+6a_3x+\sum_{n\ge4}n(n-1)a_nx^{n-2}

\displaystyle\sum_{n\ge0}a_nx^{n+1}=a_0x+\sum_{n\ge1}a_nx^{n+1}

Re-index the first sum to have it start at n=1 (to match the the other two sums):

\displaystyle\sum_{n\ge4}n(n-1)a_nx^{n-2}=\sum_{n\ge1}(n+3)(n+2)a_{n+3}x^{n+1}

So now the ODE is

\displaystyle\left(2a_2+6a_3x+\sum_{n\ge1}(n+3)(n+2)a_{n+3}x^{n+1}\right)+\sum_{n\ge1}na_nx^{n+1}-\left(a_0x+\sum_{n\ge1}a_nx^{n+1}\right)=0

Consolidate into one series starting n=1:

\displaystyle2a_2+(6a_3-a_0)x+\sum_{n\ge1}\bigg[(n+3)(n+2)a_{n+3}+(n-1)a_n\bigg]x^{n+1}=0

Suppose we're given initial conditions y(0)=a_0 and y'(0)=a_1 (which follow from setting x=0 in the power series representations for y and y', respectively). From the above equation it follows that

\begin{cases}2a_2=0\\6a_3-a_0=0\\(n+3)(n+2)a_{n+3}+(n-1)a_n=0&\text{for }n\ge2\end{cases}

Let's first consider what happens when n=3k-2, i.e. n\in\{1,4,7,10,\ldots\}. The recurrence relation tells us that

a_4=-\dfrac{1-1}{(1+3)(1+2)}a_1=0\implies a_7=0\implies a_{10}=0

and so on, so that a_{3k-2}=0 except for when k=1.

Now let's consider n=3k-1, or n\in\{2,5,8,11,\ldots\}. We know that a_2=0, and from the recurrence it follows that a_{3k-1}=0 for all k.

Finally, take n=3k, or n\in\{0,3,6,9,\ldots\}. We have a solution for a_3 in terms of a_0, so the next few terms (k=2,3,4) according to the recurrence would be

a_6=-\dfrac2{6\cdot5}a_3=-\dfrac2{6\cdot5\cdot3\cdot2}a_0=-\dfrac{a_0}{6\cdot3\cdot5}
a_9=-\dfrac5{9\cdot8}a_6=\dfrac{a_0}{9\cdot6\cdot3\cdot8}
a_{12}=-\dfrac8{12\cdot11}a_9=-\dfrac{a_0}{12\cdot9\cdot6\cdot3\cdot11}

and so on. The reordering of the product in the denominator is intentionally done to make the pattern clearer. We can surmise the general pattern for n=3k as

a_{3k}=\dfrac{(-1)^{k+1}a_0}{(3k\cdot(3k-3)\cdot(3k-2)\cdot\cdots\cdot6\cdot3\cdot(3k-1)}
a_{3k}=\dfrac{(-1)^{k+1}a_0}{3^k(k\cdot(k-1)\cdot\cdots\cdot2\cdot1)\cdot(3k-1)}
a_{3k}=\dfrac{(-1)^{k+1}a_0}{3^kk!(3k-1)}

So the series solution to the ODE is given by

y=\displaystyle\sum_{n\ge0}a_nx^n
y=a_1x+\displaystyle\sum_{k\ge0}\frac{(-1)^{k+1}a_0}{3^kk!(3k-1)}

Attached is a plot of a numerical solution (blue) to the ODE with initial conditions sampled at a_0=y(0)=1 and a_1=y'(0)=2 overlaid with the series solution (orange) with n=3 and n=6. (Note the rapid convergence.)

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Which of the following best completes the proof showing that ΔWXZ ~ ΔXYZ? triangles WXZ and XYZ that share side XZ with right an
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Answer:

81

Step-by-step explanation:

6 over 12 equals 3 over 6 6 over 3 equals 6 over 12 6 over 6 equals 3 over 12 12 over 6 equals 3 over 6

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