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3 years ago

13

A principle of $4570 is placed into account that earns 4.5 interest if the interest is compounded annually how much money will b

e in the account

1 answer:

Lelu [443]3 years ago

3 0

Answer:

205.65 interest

Step-by-step explanation:

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the graph shows that lisa's earnings are proportional to the number of hours that she works.This relationship can be represented

Annette [7]

c. 25 because she gets $25 per hour

3 0

3 years ago

Two lines, A and B, are represented by the following equations:

ch4aika [34]

Line A: 2x+2y=8 ⇒2y=-2x+8 ⇒ y=-x+4
Line B: x+y=3 ⇒y=-x+3

Therefore: y=y then:
-x+4=-x+3
-x+x=3-4
0≠-1 ⇒There is no solution because the lines are parallels

Answer: There is no solution

8 0

3 years ago

Read 2 more answers

. (08.01 MC) 1. Find the volume of a cylinder with a diameter of 8 inches and a height that is three times the radius. Use 3.14

icang [17]

Answer:

602.88 in³

12

Step-by-step explanation:

Formula for volume of a cylinder = πr² · h

radius (r) = 1/2(diameter)

1. Set up the equation

radius = 4

(3.14)(4²)(3·4)

2. Solve

3.14(16)(12) = 602.88 in³

Formula for volume of a cone = 1/3πr² · h

The formula of a cone is 1/3 the volume of a cylinder. Therefore, a cone that fits perfectly within the dimensions of a cylinder would have a volume equal to 1/3 of the volume of the cylinder.

1. Set up the equation and solve

36 ÷ 3 = 12

6 0

3 years ago

PLEASE HELP MEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEE

timama [110]

Answer:

add 6 to both sides and dividing both sides by 5.

3 0

2 years ago

Read 2 more answers

Find a particular solution to the differential equation y′′+2y′+y=2t2−t+3e−2t. y′′+2y′+y=2t2−t+3e−2t.

Jlenok [28]

$y''+2y'+y=2t^2-t+3e^{-2t}$

The characteristic equation is

$r^2+2r+1=(r+1)^2=0\implies r=-1$

so the characteristic solution is

$y_c=C_1e^{-t}+C_2te^{-t}$

For the particular solution, we can try looking for a solution of the form

$y_p=at^2+bt+c+de^{-2t}$
$\implies{y_p}'=2at+b-2de^{-2t}$
$\implies{y_p}''=2a+4de^{-2t}$

Substituting into the ODE, we have

$(2a+4de^{-2t})+2(2at+b-2de^{-2t})+(at^2+bt+c+de^{-2t})=2t^2-t+3e^{-2t}$
$at^2+(4a+b)t+(2a+2b+c)+de^{-2t}=2t^2-t+3e^{-2t}$
$\implies\begin{cases}a=2\\4a+b=-1\\2a+2b+c=0\\d=3\end{cases}\implies a=2,b=-9,c=14,d=3$

So the general solution to the ODE is

$y=y_c+y_p$
$y=C_1e^{-t}+C_2te^{-t}+2t^2-9t+14+3e^{-2t}$

4 0

3 years ago