Combine the like terms to create an equivalent expression:

2. Harry just deposited $1500 into a savings account giving 6% interest compounded quarterly.a) How much will be in the account

Studentka2010 [4]

The formula for compound interest is:

$\begin{gathered} A=P(1+\frac{r}{n})^{nt} \\ \text{where,} \\ A=\text{ Final amount} \\ r=\text{ Interest rate} \\ n=\text{ Number of times interest applied per period} \\ t=\text{ Number of time period elapsed} \\ P=\text{ Intial principal balance} \end{gathered}$

Given data:

$\begin{gathered} P=\text{ \$1500} \\ r=6\text{ \%}=0.06 \\ n=4\text{ times (compounded quarterly)} \end{gathered}$

a. After ten years, that is t = 10 years, the amount in the account will be

$\begin{gathered} A=1500(1+\frac{0.06}{4})^{4\times10} \\ A=\text{ }1500(1+0.015)^{40} \\ A=\text{ }1500(1.015)^{40} \\ A=\text{ \$2721.03} \end{gathered}$

b. After twenty years, that is t = 20 years, the amount in the account will be:

$\begin{gathered} A=1500(1+\frac{0.06}{4})^{4\times20} \\ A=1500(1.015)^{4\times20} \\ A=1500(1.015)^{80} \\ A=\text{ \$}4935.99 \end{gathered}$

c. The time it takes for Harry's initial account value to double will be:

$\begin{gathered} A=2\text{ x initial value = 2 }\times\text{ \$1500 = \$3000} \\ 3000=1500(1.015)^{4t} \\ (1.015)^{4t}=\frac{3000}{1500} \\ (1.015)^{4t}=2 \\ \text{ Find the logarithm of both sides} \\ \ln (1.015)^{4t}=\ln 2 \\ 4t=\frac{\ln 2}{\ln 1.015} \\ 4t=46.56 \\ t=\frac{46.56}{4}=11.64 \end{gathered}$

Therefore, the time it takes Harry's initial account to double is approximately 11 years

8 0

1 year ago

g A cylindrical tank with radius 7 m is being filled with water at a rate of 6 mଷ/min. How fast is the height of the water incre

RoseWind [281]

Answer:

6/(49π) ≈ 0.03898 m/min

Step-by-step explanation:

V = πr²h . . . . formula for the volume of a cylinder

dV/dt = πr²·dh/dt . . . . differentiate to find rate of change

Solving for dh/dt and filling in the numbers, we have ...

dh/dt = (dV/dt)/(πr²) = (6 m³/min)/(π(7 m)²) = 6/(49π) m/min

dh/dt ≈ 0.03898 m/min

8 0

3 years ago

Find the the perimeter of triangle JKL

Leona [35]

Answer:

60

Step-by-step explanation:

Tangents to a circle from a common external point are congruent, thus

JA = JB = 6

AL = KC = 11

CK = KB = 13

Thus

perimeter = 2(6) + 2(11) + 2(13) = 12 + 22 + 26 = 60

7 0

3 years ago

• 6x – 15y = 15<br> y=2 x - 1<br> 5

yuradex [85]

Answer:

Slope=

2.000

0.800

=0.400

x−intercept=

2 /5

=2.50000

y−intercept=

−5 /5

=

−1

1

=−1.00000

Step-by-step explanation:

STEP

1

:

Pulling out like terms

1.1 Pull out like factors :

6x - 15y - 15 = 3 • (2x - 5y - 5)

Equation at the end of step

1

:

STEP

2

:

Equations which are never true

2.1 Solve : 3 = 0

This equation has no solution.

A a non-zero constant never equals zero.

Equation of a Straight Line

2.2 Solve 2x-5y-5 = 0

Tiger recognizes that we have here an equation of a straight line. Such an equation is usually written y=mx+b ("y=mx+c" in the UK).

"y=mx+b" is the formula of a straight line drawn on Cartesian coordinate system in which "y" is the vertical axis and "x" the horizontal axis.

In this formula :

y tells us how far up the line goes

x tells us how far along

m is the Slope or Gradient i.e. how steep the line is

b is the Y-intercept i.e. where the line crosses the Y axis

The X and Y intercepts and the Slope are called the line properties. We shall now graph the line 2x-5y-5 = 0 and calculate its properties

8 0

3 years ago

<img src="https://tex.z-dn.net/?f=%5C%5C%20%5Cleq%20%5Cbeta%20%5Calpha%20%5Calpha" id="TexFormula1" title="\\ \leq \beta \alpha

nadya68 [22]

Hi lol I don’t understand you’re question I need more context

8 0

3 years ago