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2 years ago

11

HELP ASAP !! Sloane kicked a soccer ball at a speed of 36 feet per second. If the ball never leaves the ground, then it can be r

epresented by
the function H(t) = - 16t2 + 36t. Determine the time the ball traveled.
t = 0.44 seconds
t = 2.25 seconds
t = 16 seconds
t = 36 seconds

2 answers:

murzikaleks [220]2 years ago

5 0

Should be 2.25 seconds

Korolek [52]2 years ago

4 0

Answer: B: 2.25 seconds

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Some friends of yours collected books for a book drive. They collected 64 Fiction, and 48 Non-Fiction books. They asked you to m

Cloud [144]

As all of the bundles have the same content inside, so assuming that there is x number of Fiction books and y number of Non-fiction books in each bundle.

Let n be the total number of bundles that my team can send.

There are 64 Fiction books, so

nx=64 ...(i)

Or x=64/n ...(ii)

also, there are 48 Non-Fiction books, so

ny=48 ...(iii)

Or y=48/n ...(iv)

Observing that the numbers x, y, and n are counting numbers and from equations (i) and (iii), n is the common factor of 64 and 48.

The possible common factors of 64 and 48 are,

n=1,2,4,8, and 16.

So, my team can send 1,2,3,4,8 or 16 bundles of books.

Now, from equations (ii) and (iv),

For n=1:

x=64/1=64

y=48/1=48

So, for 1 bundle the number of Fiction and Non-fictions books are 64 and 48 respectively.

For n=2:

x=64/2=32

y=48/2=48

So, for 2 bundles, the number of Fiction and Non-fictions books are 32 and 24 respectively.

For n=4:

x=64/4=16

y=48/4=12

So, for 4 bundles, the number of Fiction and Non-fictions books are 16 and 12 respectively.

For n=8:

x=64/8=8

y=48/8=6

So, for 8 bundles, the number of Fiction and Non-fictions books are 8 and 6 respectively.

For n=16:

x=64/16=4

y=48/16=3

So, for 16 bundles, the number of Fiction and Non-fictions books are 4 and 3 respectively.

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3 years ago

Can I please have some HELP!!! (Brainllest)

Solnce55 [7]

Answer:

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3 years ago

Solution for dy/dx+xsin 2 y=x^3 cos^2y

vichka [17]

Rearrange the ODE as

$\dfrac{\mathrm dy}{\mathrm dx}+x\sin2y=x^3\cos^2y$
$\sec^2y\dfrac{\mathrm dy}{\mathrm dx}+x\sin2y\sec^2y=x^3$

Take $u=\tan y$ , so that $\dfrac{\mathrm du}{\mathrm dx}=\sec^2y\dfrac{\mathrm dy}{\mathrm dx}$ .

Supposing that $|y|$ , we have $\tan^{-1}u=y$ , from which it follows that

$\sin2y=2\sin y\cos y=2\dfrac u{\sqrt{u^2+1}}\dfrac1{\sqrt{u^2+1}}=\dfrac{2u}{u^2+1}$
$\sec^2y=1+\tan^2y=1+u^2$

So we can write the ODE as

$\dfrac{\mathrm du}{\mathrm dx}+2xu=x^3$

which is linear in $u$ . Multiplying both sides by $e^{x^2}$ , we have

$e^{x^2}\dfrac{\mathrm du}{\mathrm dx}+2xe^{x^2}u=x^3e^{x^2}$
$\dfrac{\mathrm d}{\mathrm dx}\bigg[e^{x^2}u\bigg]=x^3e^{x^2}$

Integrate both sides with respect to $x$ :

$\displaystyle\int\frac{\mathrm d}{\mathrm dx}\bigg[e^{x^2}u\bigg]\,\mathrm dx=\int x^3e^{x^2}\,\mathrm dx$
$e^{x^2}u=\displaystyle\int x^3e^{x^2}\,\mathrm dx$

Substitute $t=x^2$ , so that $\mathrm dt=2x\,\mathrm dx$ . Then

$\displaystyle\int x^3e^{x^2}\,\mathrm dx=\frac12\int 2xx^2e^{x^2}\,\mathrm dx=\frac12\int te^t\,\mathrm dt$

Integrate the right hand side by parts using

$f=t\implies\mathrm df=\mathrm dt$
$\mathrm dg=e^t\,\mathrm dt\implies g=e^t$
$\displaystyle\frac12\int te^t\,\mathrm dt=\frac12\left(te^t-\int e^t\,\mathrm dt\right)$

You should end up with

$e^{x^2}u=\dfrac12e^{x^2}(x^2-1)+C$
$u=\dfrac{x^2-1}2+Ce^{-x^2}$
$\tan y=\dfrac{x^2-1}2+Ce^{-x^2}$

and provided that we restrict $|y|$ , we can write

$y=\tan^{-1}\left(\dfrac{x^2-1}2+Ce^{-x^2}\right)$

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3 years ago

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Furkat [3]

Answer:

3.1782437e+42

Step-by-step explanation:

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3 years ago

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Is the table a function?

Dmitry_Shevchenko [17]

Answer:

short answer no

Step-by-step explanation:

there's repeated values for y

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3 years ago