Ask question

Ask question

All categories

3 years ago

14

Jordan wants to know how long it will take for the money she deposited to double. she has an interest rate of 4 percent. calcula

te how long it will take her money to double. four years sixteen years eighteen years eighteen months

2 answers:

Sholpan [36]3 years ago

7 0

18 months would be the answer.

hodyreva [135]3 years ago

5 0

2=1.04^t

t=ln2/(ln1.04)

t=17.673 (approx)

so t=18 (although you will have slightly more than double)

You might be interested in

If a piece of ribbon is 5 and 1 over 2. Feet long, how long is it in inches? 33 inches 55 inches 60 inches 66 inches

galben [10]

Answer:The answeris 66 inches.

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

I need help with this please it’s some this about diameter idk

lianna [129]

120 cubic feet I think

7 0

3 years ago

Read 2 more answers

Which function below is the inverse of f(x) = The quantity of three x plus five, over seven?

Nastasia [14]

Given function: f(x) = (3x+5)/7

y = f(x) so it's the same as y = (3x+5)/7

Swap every copy of x and y to get x = (3y+5)/7

Solve for y to get....
x = (3y+5)/7
7x = 7*(3y+5)/7
7x = 3y+5
7x-5 = 3y+5-5
7x-5 = 3y
3y = 7x-5
3y/3 = (7x-5)/3
y = (7x-5)/3

So the inverse function is f^(-1)(x) = (7x-5)/3

Answer is choice A

5 0

3 years ago

Read 2 more answers

Find the simple interest on a $2,219.00 principal, deposited for 6 years at a rate of 1.91%.

BARSIC [14]

Answer:

7000

Step-by-step explanation:

nhi batunga

6 0

3 years ago

The earth rotates once per day about an axis passing through the north and south poles, an axis that is perpendicular to the pla

forsale [732]

Answer:

a)

Speed at Equator = 463.97 meters per second

Centripetal Acceleration at Equator = $3.37*10^{-2}$ meters per second squared

b)

Speed at 30 degrees north of equator = 401.79 meters per second

Centripetal Acceleration at 30 degrees north of equator = $2.92*10^{-2}$ meters per second squared

Step-by-step explanation:

The formula is:

$v=\frac{2 \pi R}{T}$

Where

v is speed

R is radius

T is time

and another formula for centripetal acceleration:

$a_c=\frac{4 \pi^{2} R}{T^2}$

Now,

a)

at equator, the radius is radius of earth (given), time in seconds is

T = 24 * 60 * 60 = 86,400

THus,

$v_E=\frac{2 \pi (6.38*10^{6}}{86,400}=463.97$

Speed at Equator = 463.97 meters per second

Centripetal Acceleration:

$a_{cE}=\frac{v_E^2}{R_E}=\frac{463.97}{6.38*10^{6}}=3.37*10^{-2}$

Centripetal Acceleration at Equator = $3.37*10^{-2}$ meters per second squared

b)

At 30.0° north of the equator:

$R_N=R_E Cos (30)= (6.38*10^6)Cos(30)=5.53*10^6$

Now,

Speed = $v_{30N}=\frac{2 \pi (5.53*10^6)}{86,400}=401.79$

Speed at 30 degrees north of equator = 401.79 meters per second

Centripetal Acceleration:

$a_{30N}=\frac{v_E^2}{R_E}=\frac{401.79}{5.53*10^6}=2.92*10^{-2}$

Centripetal Acceleration at 30 degrees north of equator = $2.92*10^{-2}$ meters per second squared

4 0

3 years ago