Ask question

Ask question

All categories

2 years ago

7

2. A pilot can travel 448 miles with the wind in the same amount of time as 368 miles against the wind. Find the speed of the wi

nd if the pilot’s speed in still air is 255 miles per hour.

2 answers:

Maslowich2 years ago

8 0

Answer:

I think you subtract both and find the thing they have in common

Step-by-step explanation:

Lady_Fox [76]2 years ago

3 0

Answer:

wind speed = 25 miles/hr

Step-by-step explanation:

let a = speed of the wind

speed with the wind = 255 + a , since the wind is not against him

speed against the wind = 255 - a

travel time is same for against and withe wind

therefore,

speed = distance/time

time = distance/time

448/255 + a = 368/255 - a

cross multiply

448(255 - a) = 368(255 + a)

114240 - 448a = 93840 + 368a

114240 - 93840 = 448a + 368a

20400 = 816a

divide both sides by 816

a = 20400/816

a = 25 miles/hr

You might be interested in

Solve the recurrence relation: hn = 5hn−1 − 6hn−2 − 4hn−3 + 8hn−4 with initial values h0 = 0, h1 = 1, h2 = 1, and h3 = 2 using (

musickatia [10]

(a) Suppose $h_n=r^n$ is a solution for this recurrence, with $r\neq0$ . Then

$r^n=5r^{n-1}-6r^{n-2}-4r^{n-3}+8r^{n-4}$
$\implies1=\dfrac5r-\dfrac6{r^2}-\dfrac4{r^3}+\dfrac8{r^4}$
$\implies r^4-5r^3+6r^2+4r-8=0$
$\implies (r-2)^3(r+1)=0\implies r=2,r=-1$

So we expect a general solution of the form

$h_n=c_1(-1)^n+(c_2+c_3n+c_4n^2)2^n$

With $h_0=0,h_1=1,h_2=1,h_3=2$ , we get four equations in four unknowns:

$\begin{cases}c_1+c_2=0\\-c_1+2c_2+2c_3+2c_4=1\\c_1+4c_2+8c_3+16c_4=1\\-c_1+8c_2+24c_3+72c_4=2\end{cases}\implies c_1=-\dfrac8{27},c_2=\dfrac8{27},c_3=\dfrac7{72},c_4=-\dfrac1{24}$

So the particular solution to the recurrence is

$h_n=-\dfrac8{27}(-1)^n+\left(\dfrac8{27}+\dfrac{7n}{72}-\dfrac{n^2}{24}\right)2^n$

(b) Let $G(x)=\displaystyle\sum_{n\ge0}h_nx^n$ be the generating function for $h_n$ . Multiply both sides of the recurrence by $x^n$ and sum over all $n\ge4$ .

$\displaystyle\sum_{n\ge4}h_nx^n=5\sum_{n\ge4}h_{n-1}x^n-6\sum_{n\ge4}h_{n-2}x^n-4\sum_{n\ge4}h_{n-3}x^n+8\sum_{n\ge4}h_{n-4}x^n$
$\displaystyle\sum_{n\ge4}h_nx^n=5x\sum_{n\ge3}h_nx^n-6x^2\sum_{n\ge2}h_nx^n-4x^3\sum_{n\ge1}h_nx^n+8x^4\sum_{n\ge0}h_nx^n$
$G(x)-h_0-h_1x-h_2x^2-h_3x^3=5x(G(x)-h_0-h_1x-h_2x^2)-6x^2(G(x)-h_0-h_1x)-4x^3(G(x)-h_0)+8x^4G(x)$
$G(x)-x-x^2-2x^3=5x(G(x)-x-x^2)-6x^2(G(x)-x)-4x^3G(x)+8x^4G(x)$
$(1-5x+6x^2+4x^3-8x^4)G(x)=x-4x^2+3x^3$
$G(x)=\dfrac{x-4x^2+3x^3}{1-5x+6x^2+4x^3-8x^4}$
$G(x)=\dfrac{17}{108}\dfrac1{1-2x}+\dfrac29\dfrac1{(1-2x)^2}-\dfrac1{12}\dfrac1{(1-2x)^3}-\dfrac8{27}\dfrac1{1+x}$

From here you would write each term as a power series (easy enough, since they're all geometric or derived from a geometric series), combine the series into one, and the solution to the recurrence will be the coefficient of $x^n$ , ideally matching the solution found in part (a).

3 0

2 years ago

Students at a private liberal arts college are classified as being freshmen, sophomores, juniors or seniors, and also according

labwork [276]

The total number of possible classifications for the students of this college is found by multiplying 4 (which is the classification for the year level:freshman, sophomore, juniou, senior) and 2 (which is the number of sexes: female and male). So 4 x 2 = 8. There are eight possible classifications, which are:
(Male, Freshman)
(Male, Sophomore)
(Male, Junior)
(Male, Senior)
(Female, Freshman)
(Female, Sophomore)
(Female, Junior)
(Female,Senior)

5 0

2 years ago

Choose all sequences of transformations that produce the same image of a given figure.

elena-14-01-66 [18.8K]

<em><u>Your answer: </u></em> a reflection across the y-axis followed by a clockwise rotation 90° about the origin. A clockwise rotation 90° about the origin followed by a reflection across the x-axis. A counter-clockwise rotation 90° about the origin followed by a reflection across the y-axis. A reflection across the x-axis followed by a counter-clockwise rotation 90° about the origin.

Hope this helps <3

Stay safe

Stay Warm

-Carrie

Ps. It would mean a lot if you marked this brainliest

5 0

2 years ago

Find the equation of the line passing through the point (2,−4) that is parallel to the line y=3x+2

Naily [24]

Hey there! :)

Line passes through (2, -4) & parallel to y = 3x+ 2

Let's start off by identifying what our slope is. In the slope-intercept form y=mx+b, we know that "m" is our slope. "M" is simply a place mat so if we look at our given line, the "m" value is 3. Therefore, our slope is 3.

We should also note that we're looking for a line that's parallel to the given one. This means that our new line has the same slope as our given line. Therefore, our slope for our new line will be 3.

Now, we use point-slope form ( y-y₁=m(x-x₁) ) to complete our task of finding a line that passes through (2, -4) with a slope of 3.

y-y₁=m(x-x₁)

Let's start by plugging in 3 for m (our slope), 2 for x1 and -4 for y1.

y - (-4) = 3(x - 2)

Simplify.

y + 4 = 3x - 6

Simplify by subtracting 4 from both sides.

y = 3x - 10

~Hope I helped!~

8 0

2 years ago

What is the approximate radius of a 112 48 cd nucleus?

Kryger [21]

What is it comparing it to?

<span>2√3 = 1.26 times as large </span>

7 0

1 year ago

Read 2 more answers