Find the percent of 54 to 90

Dwight deposits $150 into his new savings account. The account earns 5% interest compounded annually.

GrogVix [38]

Answer:

$A=\$150(1.05)^{t}$

Step-by-step explanation:

we know that

The compound interest formula for this problem is equal to

$A=P(1+r)^{t}$

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 in years

in this problem we have

$P=\$150\\ r=5\%=0.05$

substitute in the formula above

$A=\$150(1+0.05)^{t}$

$A=\$150(1.05)^{t}$

8 0

3 years ago

What is a 15 yard gain

Alex_Xolod [135]

Are you by any chance referring to football? If you are, it is either a play in which by running or passing the offense moves down the field towards the opponent's endzone 15 yards, or it can be obtained by the defense committing an egregious penalty, such as a facemask, unsportsmanlike conduct, or unnecessary roughness.

7 0

3 years ago

What is the rate of change

kow [346]

The average rate of change between two input values is the total change of the function values (output values) divided by the change in the input values.

3 0

2 years ago

Read 2 more answers

Find the center and radius for the circle by the equations: x2 + y2 + 5x - y + 2 = 0.

lord [1]

X2+5x+y2-y=-2
X2+2*5x2+(5/2)^2-(5/2)^2+y2-2*y/2+(1/2)^2-(1/2)^2=-2
(x+5/2)^2+(y-1/2)^2-13/2=-2
(x+5/2)^2+(y-1/2)^2=9/2
So centre =(-5/2,1/2)
Radius=(9/2)^(1/2)

7 0

3 years ago

Suppose that a large mixing tank initially holds 500 gallons of water in which 50 pounds of salt have been dissolved. Another br

Lina20 [59]

Answer:

The differential equation for the amount of salt A(t) in the tank at a time t > 0 is $\frac{dA}{dt}=12 - \frac{2A(t)}{500+t}$ .

Step-by-step explanation:

We are given that a large mixing tank initially holds 500 gallons of water in which 50 pounds of salt have been dissolved. Another brine solution is pumped into the tank at a rate of 3 gal/min, and when the solution is well stirred, it is then pumped out at a slower rate of 2 gal/min.

The concentration of the solution entering is 4 lb/gal.

Firstly, as we know that the rate of change in the amount of salt with respect to time is given by;

$\frac{dA}{dt}= \text{R}_i_n - \text{R}_o_u_t$

where, $\text{R}_i_n$ = concentration of salt in the inflow $\times$ input rate of brine solution

and $\text{R}_o_u_t$ = concentration of salt in the outflow $\times$ outflow rate of brine solution

So, $\text{R}_i_n$ = 4 lb/gal $\times$ 3 gal/min = 12 lb/gal

Now, the rate of accumulation = Rate of input of solution - Rate of output of solution

= 3 gal/min - 2 gal/min

= 1 gal/min.

It is stated that a large mixing tank initially holds 500 gallons of water, so after t minutes it will hold (500 + t) gallons in the tank.

So, $\text{R}_o_u_t$ = concentration of salt in the outflow $\times$ outflow rate of brine solution

= $\frac{A(t)}{500+t} \text{ lb/gal } \times 2 \text{ gal/min}$ = $\frac{2A(t)}{500+t} \text{ lb/min }$

Now, the differential equation for the amount of salt A(t) in the tank at a time t > 0 is given by;

= $\frac{dA}{dt}=12\text{ lb/min } - \frac{2A(t)}{500+t} \text{ lb/min }$

or $\frac{dA}{dt}=12 - \frac{2A(t)}{500+t}$ .

4 0

3 years ago