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How much more would $1,000 earn in 5 years in an account compounded continuously than an account compounded quarterly if the int

s2008m [1.1K]

Answer: There is a difference of $ 1.0228.

Explanation: Given, initial amount or principal = $ 1000,

Time= 5 years and given compound rate of interest = $3.7%

Now, Since the amount in compound continuously,

$A= Pe^{rt}$ , where, r is the rate of compound interest, P is the principal amount and t is the time.

Here, P=$ 1000, t=5 years and r= $3.7%,

Thus, amount in compound continuously , $A=1000e^{3.7\times5/100}$

⇒ $A=1000e^{18.5}=1000\times 1.20321844013=1203.21844013$

Therefore, interest in this compound continuously rate =1203.21844013-1000=203.21844013

now, Since the amount in compound quarterly,

$A=P(1+\frac{r/4}{100} )^{4t}$ , where, r is the rate of compound interest, P is the principal amount and t is the time.

Thus, amount in compound quarterly, $A=1000(1+\frac{3.7/4}{100} )^{4\times5}$

⇒ $A=1000(1+\frac{3.7}{400} )^{20}$

⇒ $A= 1202.19567617$

Therefore, interest in this compound quarterly rate=1202.19567617-1000=202.19567617

So, the difference in these interests=203.21844013-202.19567617=1.02276396 ≈1.0228

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2 years ago

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What is the equation of the line that is perpendicular to y= -3x + 1 and passes through (2,3)?

Aleonysh [2.5K]

Answer:

$\large\boxed{y=\dfrac{1}{3}x+2\dfrac{1}{3}}$

Step-by-step explanation:

$\text{Let}\ k:y=_1x+b_1\ \text{and}\ l:y=m_2x+b_2.\\\\l\ \perp\ k\iff m_1m_2=-1\to m_2=-\dfrac{1}{m_1}\\============================\\\\\text{We have}\ y=-3x+1\to m_1=-3.\\\\\text{Therefore}\ m_2=-\dfrac{1}{-3}=\dfrac{1}{3}.\\\\\text{The equation of the searched line:}\ y=\dfrac{1}{3}x+b.\\\\\text{The line passes through }(2,\ 3).$

$\text{Put the coordinates of the point to the equation.}\ x=2,\ y=3:\\\\3=\dfrac{1}{3}(2)+b\\\\3=\dfrac{2}{3}+b\qquad\text{subtract}\ \dfrac{2}{3}\ \text{from both sides}\\\\b=2\dfrac{1}{3}$