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miss Akunina [59]
3 years ago
11

b. Write a complete program for the following situation related to setting the speed of a car to preset values before starting a

journey: • If speed is less than lower_ set_ speed, display the message "accelerate" to the screen, increase the car's speed to the lower set speed, and output the new speed to the screen. • If speed is greater than upper_set_speed, print the message "apply brake" to the screen, decrease the car's speed to the upper set speed, and output the new speed to the screen. • Give all necessary documentation in the comments, including the file name. • Allow the program to exit at the end. • Output the speed to the screen. • How many different initial speeds will you need to test this code?
Mathematics
1 answer:
Oksi-84 [34.3K]3 years ago
4 0
Answer:4

Explanation:
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Stella [2.4K]

Answer:

There are all true statements

Step-by-step explanation:

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6 0
2 years ago
y′′ −y = 0, x0 = 0 Seek power series solutions of the given differential equation about the given point x 0; find the recurrence
sukhopar [10]

Let

\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = a_0 + a_1x + a_2x^2 + \cdots

Differentiating twice gives

\displaystyle y'(x) = \sum_{n=1}^\infty na_nx^{n-1} = \sum_{n=0}^\infty (n+1) a_{n+1} x^n = a_1 + 2a_2x + 3a_3x^2 + \cdots

\displaystyle y''(x) = \sum_{n=2}^\infty n (n-1) a_nx^{n-2} = \sum_{n=0}^\infty (n+2) (n+1) a_{n+2} x^n

When x = 0, we observe that y(0) = a₀ and y'(0) = a₁ can act as initial conditions.

Substitute these into the given differential equation:

\displaystyle \sum_{n=0}^\infty (n+2)(n+1) a_{n+2} x^n - \sum_{n=0}^\infty a_nx^n = 0

\displaystyle \sum_{n=0}^\infty \bigg((n+2)(n+1) a_{n+2} - a_n\bigg) x^n = 0

Then the coefficients in the power series solution are governed by the recurrence relation,

\begin{cases}a_0 = y(0) \\ a_1 = y'(0) \\\\ a_{n+2} = \dfrac{a_n}{(n+2)(n+1)} & \text{for }n\ge0\end{cases}

Since the n-th coefficient depends on the (n - 2)-th coefficient, we split n into two cases.

• If n is even, then n = 2k for some integer k ≥ 0. Then

k=0 \implies n=0 \implies a_0 = a_0

k=1 \implies n=2 \implies a_2 = \dfrac{a_0}{2\cdot1}

k=2 \implies n=4 \implies a_4 = \dfrac{a_2}{4\cdot3} = \dfrac{a_0}{4\cdot3\cdot2\cdot1}

k=3 \implies n=6 \implies a_6 = \dfrac{a_4}{6\cdot5} = \dfrac{a_0}{6\cdot5\cdot4\cdot3\cdot2\cdot1}

It should be easy enough to see that

a_{n=2k} = \dfrac{a_0}{(2k)!}

• If n is odd, then n = 2k + 1 for some k ≥ 0. Then

k = 0 \implies n=1 \implies a_1 = a_1

k = 1 \implies n=3 \implies a_3 = \dfrac{a_1}{3\cdot2}

k = 2 \implies n=5 \implies a_5 = \dfrac{a_3}{5\cdot4} = \dfrac{a_1}{5\cdot4\cdot3\cdot2}

k=3 \implies n=7 \implies a_7=\dfrac{a_5}{7\cdot6} = \dfrac{a_1}{7\cdot6\cdot5\cdot4\cdot3\cdot2}

so that

a_{n=2k+1} = \dfrac{a_1}{(2k+1)!}

So, the overall series solution is

\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = \sum_{k=0}^\infty \left(a_{2k}x^{2k} + a_{2k+1}x^{2k+1}\right)

\boxed{\displaystyle y(x) = a_0 \sum_{k=0}^\infty \frac{x^{2k}}{(2k)!} + a_1 \sum_{k=0}^\infty \frac{x^{2k+1}}{(2k+1)!}}

4 0
3 years ago
Solve for z −cz+6z=tz+83
Alborosie

Answer: z=83/c - 7 + t

Step-by-step explanation:

6 0
3 years ago
A cube has a side length of 7 cm a cylinder has a radius of 5 cm and a height of 4 cm
densk [106]

Answer:

Cube has larger volume than cylinder.

Step-by-step explanation:

Given:

Side of a cube = 7 cm

Radius of a cylinder (r) = 5 cm

Height of a  (h) = 4 cm  

Find:

Larger volume = ?

Computation:

Volume of cube = Side³

Volume of cube = 7³

Volume of cube = 343 cm³

Volume\ of\  cylinder=\pi r^2h\\\\Volume\ of\  cylinder=\frac{22}{7} (5)^2(4)\\\\Volume\ of\  cylinder= 314.28

Cube has larger volume than cylinder.

5 0
3 years ago
What would be the best estimate for the height of a front door?
solmaris [256]

Answer: The height of a standard door is about 2 m. The length of an adult pace is about 1 m. The length of a size 8 shoe is about 30 cm. Most adults are between 1.5 m and 1.8 m in height.

Step-by-step explanation:

5 0
3 years ago
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