Mathematics, 12.02.2020 05:47 pjgolden04

B) Alex invests $2000 in an account that has a 6% annual rate of growth. To the nearest year, when
will the investment be worth $3600?

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Answer from: firdausmohammed80

part 1) 10 years

part 2) 10 years

Step-by-step explanation:

The correct question is:

Part 1) Alex invests $2000 in an account that has a 6% annual rate of growth compounded annually. To the nearest year, when will the investment be worth $3600?

Part 2) Alex invests $2000 in an account that has a 6% annual rate of growth compounded continuously. To the nearest year, when will the investment be worth $3600?

Part 1) we know that

The compound interest formula is equal to

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

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest in decimal

t is Number of Time Periods

n is the number of times interest is compounded per year

in this problem we have

$P=\$2,000\\A=\$3,600\\ r=6\%=6/100=0.06\\n=1$

substitute in the formula above

$3,600=2,000(1+\frac{0.06}{1})^{t}$

$1.8=(1.06)^{t}$

Apply log both sides

$log(1.8)=log[(1.06)^{t}]$

Applying property of exponents

log(1.8)=(t)log(1.06)

t=log(1.8)/log(1.06)

$t=10.09\ years$

Round to the nearest year

$t=10\ years$

Part 2) we know that

The formula to calculate continuously compounded interest is equal to

$A=P(e)^{rt}$

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest in decimal

t is Number of Time Periods

e is the mathematical constant number

we have

$P=\$2,000\\A=\$3,600\\ r=6\%=6/100=0.06$

substitute in the formula above

$3,600=2,000(e)^{0.06t}$

$1.8=[e]^{0.06t}$

Apply ln both sides

$ln(1.8)=ln[e]^{0.06t}$

ln(1.8)=(0.06t)ln[e]

t=ln(1.8)/0.06

$t=9.80\ years$

Round to the nearest year

$t=10\ years$

Answer from: Quest

step-by-step explanation:

Answer from: Quest

la parábola es el lugar

geométrico en el cual los puntos tienen una distancia idéntica, equidistantes,

de un punto denominado foco, f, y una recta denominada directriz

el radio vector de una

parábola es el segmento que existe entre un punto cualquiera y el foco.

por la ecuación se observa

que la parábola abre hacia la derecha, por lo cual podemos usar las siguientes fórmulas

la ecuación ordinaria de la

parábola es y^2 = 4px, donde p es la longitud desde el vértice de la parábola al

foco.

la fórmula de la directriz

es x + p = 0

resolviendo:

y^2-9x = 0

y^2 = 9x

4p = 9, entonces p = 9/4

entonces el foco tendrá la

coordenada f(9/4,0)

la directriz vendrá

expresada por

x + p = 0

x = -p

x = -9/4

el problema indica que la

ordenada es igual a 6, entonces

(6)^2=9x

x = 36/9

x = 4

entonces el punto al cual

hay que obtener la distancia será p(4,6)

calculando se tendrá: pf =

pf = 6,25 u

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step-by-step explanation:

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