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user100 [1]
2 years ago
12

Select the correct answer.

Mathematics
2 answers:
Mariulka [41]2 years ago
7 0

Answer: KE=\frac{1}{2} mv^{2}

Step-by-step explanation:

- the correct equation is: KE=\frac{1}{2} mv^{2}

- KE is kinetic energy

- m is mass

- v is velocity

besides that, i don't really know how to explain it but hope this helps :)

Travka [436]2 years ago
3 0

Answer:

KE = 1/2 mv²

In terms of Mass Velocity

KE = mv²

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What is the domain and derivative of f(x)=ln(2xsqrt(2+x))?
harkovskaia [24]
Domain is the numbers yo can use
we has a sqrt and a ln
we cant have any neagtive ln's or sqrts
x cannot be negative
it cannot be 0 either
domain is all numbers bigger than 0


deritivive
apply chain rule
dy/dx(f(g(x))=f'(g(x))g'(x)
and also
dy/dx (lnx)=1/x
and
dy/dx(f(x)g(x))=f'(x)g(x)+g'(x)f(x)
and from experience
dy/dx(sqrtx)=1/(2√x)


so

dy/dx(ln(2xsqrt(2+x)))=
(1/(2xsqrt(2+x))(2)(sqrt(2+x)+(x/(2sqrt(2+x))=
(3x+4)/(2x(x+2))



domain is all real numbers greater than 0
derivitive is \frac{3x+4}{2x(x+2)}
7 0
3 years ago
Write an equation to model growth or decay in the following scenario. A stock is declining at a rate of 25% of its value every 2
yawa3891 [41]

The exponential function that models this situation is:

y(t) = 225(0.75)^{\frac{t}{2}}

------------------------------------------

A decaying exponential function has the following format:

y(t) = y(0)(1 - r)^t

In which:

  • y(0) is the initial value.
  • r is the decay rate, as a decimal.

In this problem:

  • The stock started at $225, thus y(0) = 225.
  • Declines at a rate of 25% every 2 weeks, thus r = 0.25, t = \frac{t}{2}

The equation is:

y(t) = y(0)(1 - r)^t

y(t) = 225(1 - 0.25)^{\frac{t}{2}}

y(t) = 225(0.75)^{\frac{t}{2}}

A similar problem is given at brainly.com/question/24282972

3 0
3 years ago
Felicia wants to tell her Italian friend, Viviana, how cold it gets in the winter where Felicia lives—down to -17°F. Viviana has
nignag [31]
1°F is equal to -17.2222°C, so -17°F times 17.2222°C would be approximately:
-27<span>°C</span>
8 0
4 years ago
What is 4.08 as a fraction number
Mariulka [41]
Simplified: 102/25
Mixed number: 4 2/25
8 0
3 years ago
Read 2 more answers
PLEASE HELP! WILL MARK BRAINLIEST!
makkiz [27]

Answer:

The maximum height a rider will experience is 55 feet.

Step-by-step explanation:

Let's start writing the function that defines the path of a seat on the new Ferris wheel. This function will depend of the variable ''t'' which is time.

X(t)=(x,y)

In which X(t) are the coordinates of the seat (the x - coordinate and the y - coordinate) that depend from time.

''x'' and ''y'' are functions that depend from the variable ''t''.

For this exercise :

X(t)=[-25sin(\frac{\pi}{30}t);-25cos(\frac{\pi}{30}t)+30]

In order to find the maximum height a rider will experience we will study the behaviour of the y - component from the function X(t).

The function to study is y(t)=-25cos(\frac{\pi}{30}t)+30

To find its maximum, we will derivate this function and equalize it to 0. Doing this, we will find the ''critical points'' from the function.

⇒ y(t)=-25cos(\frac{\pi}{30}t)+30  ⇒

y'(t)=\frac{5}{6}\pi sin(\frac{\pi}{30}t)

Now we equalize y'(t) to 0 ⇒

y'(t)=0 ⇒ \frac{5}{6}\pi sin(\frac{\pi}{30}t)=0

In this case it is easier to look for the values of ''t'' that verify :

sin(\frac{\pi}{30}t)=0

Now we need to find the values of ''t''. We know that :

sin(0)=0\\\\sin(\pi)=0\\sin(-\pi)=0

Therefore we can write the following equivalent equations :

\frac{\pi}{30}t=0 (I)

\frac{\pi}{30}t=\pi (II)

\frac{\pi}{30}t=-\pi (III)

From (I) we obtain t_{1}=0

From (II) we obtain t_{2}=30

And finally from (III) we obtain t_{3}=-30

We found the three critical points of y(t). To see if they are either maximum or minimum we will use the second derivative test. Let's calculate the second derivate of y(t) :

y'(t)=\frac{5}{6}\pi sin(\frac{\pi}{30}t) ⇒

y''(t)=\frac{\pi ^{2}}{36}cos(\frac{\pi }{30}t)

Now given that we have an arbitrary critical point ''t_{n}'' ⇒

If y''(t_{n})>0  then we will have a minimun at t_{n}

If y''(t_{n}) then we will have a maximum at t_{n}

Using the second derivative test with t_{1},t_{2} and t_{3} ⇒

y''(t_{1})=y''(0)=\frac{\pi ^{2}}{36} >0 ⇒ We have a minimum for t_{1}=0

y''(t_{2})=y''(30)=\frac{-\pi^{2}}{36} ⇒ We have a maximum for t_{2}=30

y''(t_{3})=y''(-30)=\frac{-\pi^{2}}{36} ⇒ We have a maximum for t_{3}=-30

The last step for this exercise will be to find the values of the maximums.

We can do this by replacing in the equation of y(t) the critical points t_{2} and t_{3} ⇒

y(t_{2})=y(30)=55

y(t_{3})=y(-30)=55

We found out that the maximum height a rider will experience is 55 feet.

3 0
3 years ago
Read 2 more answers
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