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3 years ago

The amount of money in a savings account after t years is represented by the function f(t)=3600(1.035)t .

Mathematics

2 answers:

Nataly_w [17]3 years ago

8 0

I think that I had this question a while back, I’m pretty sure the answer is C

timama [110]3 years ago

5 0

Answer: The initial amount in the account = $3600

Step-by-step explanation:

Since we have given that

The amount of money in a savings account after t years is represented by the function

$f(t)=3600(1.035)^t$

As we know the formula "Compound Interest" :

$Amount=P(1+\frac{r}{100})^t\\\\Here,\\\\\text{ P denotes principal}\\\\\text{ R denotes Rate of interest}\\\\\text{ T denotes time}$

So, According to our question,

Rate of interest = 0.35 = 35%

So, equate the both the equations , we get that

Hence, The initial amount in the account = $3600

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3 years ago

Find b and s. At The present time Mrs. B's age is six years more than four times her sons age. Three years ago she was seven tim

padilas [110]

Answer:

Mrs. B's age = y = 38 years

her son's age = x = 8 years

Step-by-step explanation:

To Find:

Mrs. B's age = y =?

her son's age = x = ?

Solution:

Let Mrs. B's age be 'y' years

and her son's age be 'x' years

Three years ago Mrs B's will be (y - 3) and her son's age will be (x - 3)

According to the first given condition,

y = 6 + 4x ...............Equation ( 1 )

According to the second given condition,

(y - 3 ) = 7 (x - 3 )..........Equation ( 2 )

equating equation 1 in equation 2 we get

6 + 4x -3 = 7x - 21

7x - 4x = 21 + 3

∴ $3x=24\\\therefore x =\frac{24}{3} \\\\\therefore x =8\ years$

Now substitute x in equation 1 we get

y = 6 + 4×8

y = 6 + 32

∴ y = 38 years

Mrs. B's age = y = 38 years

her son's age = x = 8 years

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3 years ago

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Marizza181 [45]

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3 years ago

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3 years ago

Read 2 more answers

Jeremy analyses one of his parachute jumps. He draws a graph showing his velocity up to the opening of his parachute. a) Estimat

jeyben [28]

Answer:

Jeremy's acceleration is about $1\,\frac{m}{s^2}$ at t =10 s

His average speed is about 44.5 m/s in this section of his jump approximating with points on the curve (under-estimate)

His average speed is about 46 m/s if using the tangent line (over estimate)

Step-by-step explanation:

Jeremy's acceleration can be estimated by the curve's derivative at that point. That is the slope of the tangent line to the velocity curve at x = 10 sec. Please see attached image where the tangent line is drawn in orange, and the points to use to calculate its slope are drawn in green.

These points are : (6, 42) and (14,50) which using the slope formula give:

$slope=\frac{y_2-y_1}{x_2-x_1}= \frac{50-42}{14-6}=\frac{8}{8} \frac{m}{s^2} = 1\,\frac{m}{s^2}$

So his acceleration at that point is about $1\,\frac{m}{s^2}$

Now, using about the same interval of x-values (from 6 to 14), the corresponding speeds are approximately: 40 (for time 6 seconds) and 49 (for time 14 seconds (look for the red dots on the attached image). Since the average velocity is given by the integral of the function between those points divided by the length of the interval where it is calculated:

$v_{average}=\frac{area}{interval\,\,length}$

and we don't have the actual velocity function to estimate the integral, we can approximate this area by that of a trapezoid that connects the red dots with the bottom of the horizontal axis (see red trapezoid in the image). Clearly from the image, this approximation would give us an under-estimate of the actual average speed.

The area of this trapezoid is: approximately:

$Trapezoid\,\, area=(49+40)\,8/2=356$

Then the average velocity estimated from it is:

$v_{average}=\frac{356}{8} \frac{m}{s} =44.5\,\frac{m}{s}$

If the area is approximated instead with the trapezoid form by the green points we used to calculate the acceleration (this would give us an over-estimate):

$Trapezoid\,\, area=(50+42)\,8/2=368$

Then the average velocity estimated from it is:

$v_{average}=\frac{368}{8} \frac{m}{s} =46\,\frac{m}{s}$

while his actual instantaneous velocity seems to be about 46 m/s from the graph

4 0

3 years ago