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irina [24]
3 years ago
15

The value, V, of Kalani’s stock investments over a time period, x, can be determined using the equation v=750(0.80)-x. What is t

he rate of increase or decrease associated with this account? 20% decrease 20% increase 25% decrease 25% increase
Mathematics
2 answers:
Snowcat [4.5K]3 years ago
6 0
The rate of increase and decrease associated with Kalani's stock investments would be a 25% per year increase.

Therefore, the correct option of all the answer choices you listed is: C.
Alex17521 [72]3 years ago
4 0

Answer:

Hence, the rate of change is:

25% increase.

Step-by-step explanation:

We are given a function V, of Kalani's stock investments over a time period,x as:

V(x)=750\times (0.80)^{-x}

Now the rate of increase or decrease in percent associated with this account can be calculated as:

\dfrac{V(x+1)-V(x)}{V(x)}\times 100

=\dfrac{750\times (0.80)^{-(x+1)}-750\times (0.80)^{-x}}{750\times (0.80)^{-x}}\times 100\\\\\\=\dfrac{750\times (0.80)^{-x}\times ((0.80)^{-1}-1)}{750\times (0.80)^{-x}}\times 100\\\\\\\\=(\dfrac{1}{0.80}-1)\times 100\\\\=\dfrac{1-0.80}{0.80}\times 100\\\\\\=\dfrac{0.20}{0.80}\times 100\\\\=\dfrac{20}{80}\times 100\\\\=\dfrac{1}{4}\times 100\\\\=25\%

Hence, the rate of increase is:

25%.

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masya89 [10]

Answer:

Question not complete,

So i will analyse the possible problem

Step-by-step explanation:

A tank contain 2200L

Volume V = 2200L

Solution of 0.06kg/L of sugar

Rate of entry i.e input

dL/dt=5L/min

Let y(t) be the amount of sugar in tank at any time.

But at the beginning there was no sugar in the tank

i.e, y(0)=0, this will be out initial value problem,

The rate of amount of sugar at anytime t is

dy/dt=input amount of sugar - output amount of sugar.

Now,

Then rate of input is

5L/min × 0.06kg/L

Then, input rate= 0.3kg/min

Output rate is

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then, output rate = y(t)/440 kg/min

Now then,

dy/dt=input rate -output rate

dy/dt=0.3-y/440

Cross multiply through by 400

400dy/dt=120-y

Using variable separation

400/(120-y) dy = dt

∫400/(120-y) dy = ∫dt

-400In(120-y)=t +C

In(120-y)=-t/400+C/400

C/400 is another constant, let say B

In(120-y)=-t/400+B

Take exponential of both side

120-y(t)=exp(-t/400+B)

120-y(t)=exp(-t/400)exp(B)

exp(B) is a constant let say C

-y(t)=Cexp(-t/400)-120

y(t)=120-Cexp(t/400)

Now, the initial condition

a. At the start the mass of sugar in the water is 0 because it is just pure water at start.

Therefore y(0)=0,

b. Applying this to y(t)

y(t)=120-Cexp(-t/400)

y=0, t=0

0=120-Cexp(0)

0=120-C

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Therefore,

y(t)=120 - 120exp(-t/400)

Let know the mass rate as t tends to infinity

At infinity

exp(-∞)=1/exp(∞)=1/∞=0

Then,

The exponential aspect tend to 0

Then, y(t)=120 as t tend to ∞

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Answer:

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Step-by-step explanation:

In order to evaluate a function at a given x-value, plug in that value into the actual function

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Step-by-step explanation:

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Answer:

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