How many ways can 11 photographs be arranged horizontally

when a plane flies with the wind, it can travel 1980 miles in 4.5 hours. when the plane flies in the opposite direction, against

Romashka-Z-Leto [24]

Velocity with wind = 1,980 / 4.5 = 440 mph
Velocity against wind =1,980 / 5.5 = 360 mph
Plane in still air - wind speed = 360
Plane in still air + wind speed =440
Adding both equations:
2*Plane in still air = 800
Plane in still air = 400 mph
Plane in still air + wind speed =440
Therefore, wind speed = 40 mph

7 0

2 years ago

The function D(t) defines a traveler's distance from home, in miles, as a function of time, in hours. D(t) = StartLayout enlarge

nalin [4]

- At 2 hours, the traveler is 725 miles from home.

- At 3 hours, the distance is constant, at 880 miles. --> TRUE.

- The total distance from home after 6 hours is 1,062.5 miles --> TRUE.

Step-by-step explanation:

The function D(t) is defined as follows:

$D(t) = 300t+125$ for $t< 2.5$

$D(t) = 880$ for $2.5 \leq 3.5$

$D(t) = 75t+612.5$ for $t\leq 6$

Where

t is the time in hours

D(t) is the distance covered, in miles, after t hours

Now let's analyze the different statements:

- The starting distance, at 0 hours, is 300 miles. --> FALSE. In fact, if we substitute t = 0 into the 1st equation, we get

$D(0) = (300)(0)+125 = 125$

So, the distance at t = 0 is 125 miles.

- At 2 hours, the traveler is 725 miles from home. --> TRUE. In fact, if we substitute t = 2 into the 1st equation,

$D(2) = (300)(2)+125 = 725$

- At 2.5 hours, the traveler is 875 miles from home. --> FALSE. In fact, for t=2.5 we have to use the 2nd equation, which states that the distance is:

D(t) = 880

So, not 875 miles.

- At 3 hours, the distance is constant, at 880 miles. --> TRUE. This is clearly visible from the 2nd equation: for t between 2.5 and 3.5 (so, in this case), the distance is

D(t) = 880

- The total distance from home after 6 hours is 1,062.5 miles --> TRUE. In fact, if we replace t = 6 into the last equation,

$D(6)) = 75(6)+612.5=1062.5$

Learn more about functions:

brainly.com/question/3511750

brainly.com/question/8243712

brainly.com/question/8307968

Learn more about distance:

brainly.com/question/3969582

#LearnwithBrainly

4 0

1 year ago

Read 2 more answers

Brahmagupta solved a quadratic equation of the form ax2 + bx = c using the formula x =, which involved only one solution. Using

weeeeeb [17]

X2+7x-8=0
product=-8 times 1 = -8
sum= 7
{-1, 8}
x2-1x+8x-8=0
x(x-1)+8(x-1)=0
x+8=0 or X-1=0
x=-8

4 0

1 year ago

Read 2 more answers

Rounded to the nearest hundredth, what is the positive solution to the quadratic equation 0=2x2+3x-8?

Basile [38]

Answer:

Step-by-step explanation:

The given quadratic equation is

2x^2+3x-8 = 0

To find the roots of the equation. We will apply the general formula for quadratic equations

x = -b ± √b^2 - 4ac]/2a

from the equation,

a = 2

b = 3

c = -8

It becomes

x = [- 3 ± √3^2 - 4(2 × -8)]/2×2

x = - 3 ± √9 - 4(- 16)]/2×2

x = [- 3 ± √9 + 64]/2×2

x = [- 3 ± √73]/4

x = [- 3 ± 8.544]/4

x = (-3 + 8.544) /4 or x = (-3 - 8.544) / 4

x = 5.544/4 or - 11.544/4

x = 1.386 or x = - 2.886

The positive solution is 1.39 rounded up to the nearest hundredth

8 0

2 years ago

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A conical pile of road salt has a diameter of 112 feet and a slant height of 65 feet. After a storm, the linear dimensions of th

QveST [7]

we know that

the volume of a cone is equal to

$V= \frac{1}{3} \pi r^{2}h$

in this problem

the radius is equal to

$r= \frac{112}{2}= 56ft$

1) Find the height of the cone before the storm

Applying the Pythagorean Theorem find the height

$h^{2} = l^{2}-r^{2}$

$l=65 ft$

$h^{2} = 65^{2}-56^{2}$

$h^{2} = 1,089$

$h=33 ft$

2) Find the volume before the storm

$V= \frac{1}{3}*\pi* 56^{2}*33$

$V=34,496\pi\ ft^{3}$

3) Find the volume after the storm

After a storm, the linear dimensions of the pile are 1/3 of the original dimensions

so

r=(56/3) ft

h=(33/3)=11 ft

$V= \frac{1}{3}*\pi* (56/3)^{2}*11$

$V= 1,277.63\pi\ ft^{3}$

4) Find how this change affect the volume of the pile

Divide the volume after the storm by the volume before the storm

$\frac{1,277.63 \pi }{34,496 \pi } = \frac{1}{27}$

therefore

the answer part a) is

The volume of the pile after the storm is $\frac{1}{27}$ times the original volume

Part b) Estimate the number of lane miles that were covered with salt

5) Find the amount of salt that was used during the storm

$=34,496 \pi - 1,277.63 \pi \\= 33.218.37 \pi \\= 104,358.59\ ft^{3}$

6) Find the pounds of road salt used

$104,358.59*80=8,348,687.2\ pounds$

7) Find the number of lane miles that were covered with salt

$8,348,687.2/350=23,853.39 \ lane\ miles$

therefore

the answer part b) is

the number of lane miles that were covered with salt is $23,853.39 \ lane\ miles$

Part c) How many lane miles can be covered with the remaining salt? Round your answer to the nearest lane mile

the remaining salt is equal to $1,277.63\pi\ ft^{3}$

$1,277.63\pi\ ft^{3}=4,013.79\ ft^{3}$

8) Find the pounds of road salt

$4,013.79*80=321,103.20\ pounds$

9) Find the number of lane miles

$321,103.20/350=917.44 \ lane\ miles$

therefore

the answer part c) is

the number of lane miles is $917 \ lane\ miles$

7 0

2 years ago