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statuscvo [17]
3 years ago
15

Ted invests $8,310 in a savings account with a fixed annual interest rate of 2% compounded continuously. How long will it take f

or the account balance to reach it take for the account balance to reach $9,751.88?
Mathematics
1 answer:
Vsevolod [243]3 years ago
8 0

Answer: It will take 8 years.

Step-by-step explanation:

Equation for interest compounded continuously:

A=Pe^{rt} , A = accumulated amount , P=Principal value , r =rate of interest , t= time.

Given: P=  $8,310 , r = 2%  , A= $9,751.88

9751.88=8310e^{0.02t}\\\\\Rightarrow\ \dfrac{9751.88}{8310}=e^{0.02t}\\\\\Rightarrow\ 1.17351143201=e^{0.02t}

Taking natural log on both sides

\ln (1.17351143201)=\ln (e^{0.02t})\\\\\Rightarrow\ 0.160000478068=0.02t\\\\\Rightarrow\ t=\dfrac{0.160000478068}{0.02}\\\\\Rightarrow\ t=8.0000239034\approx8

Hence, it will take 8 years.

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16x^2+?x+36 perfect square trinomial
Hunter-Best [27]

Answer:

  16x² +48x +36

Step-by-step explanation:

A perfect square trinomial is of the form ...

  (a +b)² = a² +2ab +b²

We want to match this form.

__

<h3>comparing terms</h3>

Comparing the known terms, we see ...

  16x² = a²   ⇒   a = 4x

  36 = b²   ⇒   b = 6

<h3>filling in the missing term</h3>

The missing term is the linear term:

  2ab = 2(4x)(6) = 48x

  ? = 48

4 0
2 years ago
Given ∆ABC≅∆PQR, mb=8v-9 , find m 31°<br> b.<br> 52°<br> c.<br> 49°<br> d.<br> 32°
Ipatiy [6.2K]
Here, 5v+6 = 8v-9
8v - 5v = 6 + 9
3v = 15
v = 15/3
v = 5

So, m<b = 5(5) + 6 = 25 + 6 = 31
As they are congruent, m<q would be same [ 31 degree ]

In short, Your Answer would be Option A

Hope this helps!
3 0
3 years ago
What is the area of polygon ABCDE?
marta [7]

Answer:86

Step-by-step explanation

140-(16+24+5+9) 140-54 = 86

6 0
3 years ago
In a certain region, about 6% of a city's population moves to the surrounding suburbs each year, and about 4% of the suburban po
Sedbober [7]

Answer:

City @ 2017 = 8,920,800

Suburbs @ 2017 = 1, 897, 200

Step-by-step explanation:

Solution:

- Let p_c be the population in the city ( in a given year ) and p_s is the population in the suburbs ( in a given year ) . The first sentence tell us that populations p_c' and p_s' for next year would be:

                                  0.94*p_c + 0.04*p_s = p_c'

                                  0.06*p_c + 0.96*p_s = p_s'

- Assuming 6% moved while remaining 94% remained settled at the time of migrations.

- The matrix representation is as follows:

                         \left[\begin{array}{cc}0.94&0.04\\0.06&0.96\end{array}\right] \left[\begin{array}{c}p_c\\p_s\end{array}\right] =  \left[\begin{array}{c}p_c'\\p_s'\end{array}\right]          

- In the sequence for where x_k denotes population of kth year and x_k+1 denotes population of x_k+1 year. We have:

                         \left[\begin{array}{cc}0.94&0.04\\0.06&0.96\end{array}\right] x_k = x_k_+_1

- Let x_o be the populations defined given as 10,000,000 and 800,000 respectively for city and suburbs. We will have a population x_1 as a vector for year 2016 as follows:

                          \left[\begin{array}{cc}0.94&0.04\\0.06&0.96\end{array}\right] x_o = x_1

- To get the population in year 2017 we will multiply the migration matrix to the population vector x_1 in 2016 to obtain x_2.

                          x_2 = \left[\begin{array}{cc}0.94&0.04\\0.06&0.96\end{array}\right]\left[\begin{array}{cc}0.94&0.04\\0.06&0.96\end{array}\right] x_o

- Where,

                         x_o =  \left[\begin{array}{c}10,000,000\\800,000\end{array}\right]

- The population in 2017 x_2 would be:

                         x_2 = \left[\begin{array}{cc}0.94&0.04\\0.06&0.96\end{array}\right]\left[\begin{array}{cc}0.94&0.04\\0.06&0.96\end{array}\right] \left[\begin{array}{c}10,000,000\\800,000\end{array}\right] \\\\\\x_2 = \left[\begin{array}{c}8,920,800\\1,879,200\end{array}\right]

5 0
4 years ago
A tank contains 1000 L of pure water. Brine that contains 0.05 kg of salt per liter of water enters the tank at a rate of 5 L/mi
Margaret [11]

Answer:

a) y(t)=0.65\frac{Kg}{min}(tmin)

b) y(40)=26Kg

Step-by-step explanation:

Data

Brine a (Ba)

V_{Ba}=5\frac{Lt}{min}\\  Concentration(Bca)=0.05\frac{Kg}{Lt}

Brine b (Bb)

V_{Bb}=10\frac{Lt}{min}\\  Concentration(Bcb)=0.04\frac{Kg}{Lt}

we have that per every minute the amount of solution that enters the tank is the same as the one that leaves the tank (15 Lt / min)

, then the amount of salt (y) left in the tank after (t) minutes: y=V_{Ba}*B_{ca}+V_{Bb}*B_{cb}=5\frac{Lt}{min}*0.05\frac{Kg}{Lt}+10\frac{Lt}{min}*0.04\frac{Kg}{Lt}=\\0.25\frac{Kg}{min}+0.4\frac{Kg}{min}=0.65\frac{Kg}{min}

Finally:

a) y(t)=0.65\frac{Kg}{min}(tmin)

b) y(40)=0.65\frac{Kg}{min}(40min)=26Kg

being y(t) the amount of salt (y) per unit of time (t)

5 0
3 years ago
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