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KonstantinChe [14]
3 years ago
14

Toni invests money into an account which pays a fixed rate of compound interest each year. The total value, ?V, of her investmen

t after t years is given by the formula: V = 1350 x 1.04^t Answer questions a & b a- How much money did Toni invest in pounds b- What rate of compound interest is paid each year
Mathematics
2 answers:
Butoxors [25]3 years ago
8 0
U sucknxksnalbsiansidnc
ELEN [110]3 years ago
3 0

Answer:

a. $ 1,350

b. 4%

Step-by-step explanation:

a. Given function that shows the total value of Toni's investment after t years,

V=1350(1.04)^t,

Initially, ( when he invested ),

t = 0,

Thus, the invested amount = 1350(1.04)^0 = 1350(1) = $ 1,350,

b. Since, the amount formula in compound interest,

A=P(1+\frac{r}{n})^{nt}

Where,

P = Initial amount,

r = annual rate per period,

n  = number of periods per period,

t = number of years,

Here, the amount is compounded annually,

i.e. n = 1,

By the given function,

V=(1+0.04)^t

By comparing,

r = 0.04 = 4%

Hence, 4% interest is paid annually.

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<h3>Step-by-step explanation:</h3>

3,200 + 560 + 16 √

(400 × 8) + (70 × 8) + (2 × 8) √

(400 × 8) × (70 × 8) × (2 × 8) X

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3,200 × 560 × 16 √

(400 + 70 + 2) × 8 √

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28,672,000 X

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The population of a community is known to increase at a rate proportional to the number of people present at time t. The initial
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Step-by-step explanation:

Let P be the population of the community

So the population of a community increase at a rate proportional to the number of people present at a time

That is

\frac{dp}{dt} \propto p\\\\\frac{dp}{dt} =kp\\\\ [k \texttt {is constant}]\\\\\frac{dp}{dt} -kp =0

Solve this equation we get

p(t)=p_0e^{kt}---(1)

where p is the present population

p₀ is the initial population

If the  initial population as doubled in 5 years

that is time t = 5 years

We get

2p_o=p_oe^{5k}\\\\e^{5k}=2

Apply In on both side to get

Ine^{5k}=In2\\\\5k=In2\\\\k=\frac{In2}{5} \\\\\therefore k=\frac{In2}{5}

Substitute k=\frac{In2}{5}  in p(t)=p_oe^{kt} to get

\large \boxed {p(t)=p_oe^{\frac{In2}{5}t }}

Given that population of a community is 9000 at 3 years

substitute t = 3 in {p(t)=p_oe^{\frac{In2}{5}t }}

p(3)=p_oe^{3 (\frac{In2}{5}) }\\\\9000=p_oe^{3 (\frac{In2}{5}) }\\\\p_o=\frac{9000}{e^{3(\frac{In2}{5} )}} \\\\=5937.8

<h3>Therefore, the initial population is 5937.8</h3>
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viktelen [127]

Answer:

<h2>B. 6</h2>

Step-by-step explanation:

\text{use}\\\\a^{-n}=\dfrac{1}{a^n}\\\\==========================\\\\b^{-3}=\dfrac{1}{216}\\\\\dfrac{1}{b^3}=\dfrac{1}{216}\qquad\text{inverse both sides}\\\\b^2=216\to b=\sqrt[3]{216}\\\\b=6\ \text{because}\ 6^3=216

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3) 10(2/5)=4

4) Point Q=4

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