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3 years ago

A falcon flew 70 miles in 3 hours. What was the speed of the falcon?

Mathematics

2 answers:

adell [148]3 years ago

8 0

Answer:

23.3333333 miles per hour

23 1/3 miles per hour

10.43093333333333 meters / second

Step-by-step explanation:

Take the distance and divide by the time to find the speed

70 miles / 3 hours

1 mile = 1609.34 meters

1 hours = 3600 seconds

70 miles /3 hours * 1 mile /1609.34 meters * 1 hours / 3600 seconds

10.43093333333333 meters / second

gulaghasi [49]3 years ago

7 0

Step-by-step explanation:

speed is normally measured in unit time meaning either per hour,second,minute etc

therefore for this question you will have to find what it travels per hour

if 70 is in three hours what about one hour

therefore

70/3=23.3'

so the speed will be 23.3miles per hour

(the apostrophe is to show recurring you can also round of and ignore it)

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G find the area of the surface over the given region. use a computer algebra system to verify your results. the sphere r(u,v) =

Svetach [21]

Presumably you should be doing this using calculus methods, namely computing the surface integral along $\mathbf r(u,v)$ .

But since $\mathbf r(u,v)$ describes a sphere, we can simply recall that the surface area of a sphere of radius $a$ is $4\pi a^2$ .

In calculus terms, we would first find an expression for the surface element, which is given by

$\displaystyle\iint_S\mathrm dS=\iint_S\left\|\frac{\partial\mathbf r}{\partial u}\times\frac{\partial\mathbf r}{\partial v}\right\|\,\mathrm du\,\mathrm dv$

$\dfrac{\partial\mathbf r}{\partial u}=a\cos u\cos v\,\mathbf i+a\cos u\sin v\,\mathbf j-a\sin u\,\mathbf k$
$\dfrac{\partial\mathbf r}{\partial v}=-a\sin u\sin v\,\mathbf i+a\sin u\cos v\,\mathbf j$
$\implies\dfrac{\partial\mathbf r}{\partial u}\times\dfrac{\partial\mathbf r}{\partial v}=a^2\sin^2u\cos v\,\mathbf i+a^2\sin^2u\sin v\,\mathbf j+a^2\sin u\cos u\,\mathbf k$
$\implies\left\|\dfrac{\partial\mathbf r}{\partial u}\times\dfrac{\partial\mathbf r}{\partial v}\right\|=a^2\sin u$

So the area of the surface is

$\displaystyle\iint_S\mathrm dS=\int_{u=0}^{u=\pi}\int_{v=0}^{v=2\pi}a^2\sin u\,\mathrm dv\,\mathrm du=2\pi a^2\int_{u=0}^{u=\pi}\sin u$
$=-2\pi a^2(\cos\pi-\cos 0)$
$=-2\pi a^2(-1-1)$
$=4\pi a^2$

as expected.

6 0

2 years ago

Can i have help with these questions

adell [148]

Answer:

(-3, 5), (-1, -1), (5, -3)

Step-by-step explanation:

Each pair of vertices can be one of the diagonals. Then the missing point will be found at the coordinates that are the sum of those, less the coordinates of the third point.

Given points are ...

A(-2, 2), B(1, 1), C(2, -2)

For AB a diagonal, D1 is ...

A+B-C = (-2+1-2, 2+1-(-2)) = (-3, 5)

For AC a diagonal, D2 is ...

A+C-B = (-2+2-1, 2-2-1) = (-1, -1)

For BC a diagonal, D3 is ...

B+C-A = (1+2-(-2), 1-2-2) = (5, -3)

_____

For a lot of parallelogram problems I find it easiest to work with the fact that the diagonals bisect each other. This means they both have the same midpoint, so for quadrilateral ABCD, we have (A+C)/2 = (B+D)/2. Multiplying this by 2 gives the equation we used above, A+C = B+D, so D=A+C-B. Remember, in ABCD, AC and BD are the diagonals.

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bixtya [17]

Answer:

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Step-by-step explanation:

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Sunny_sXe [5.5K]

Answer:

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