Baby Cake Shoppe FAQ’s

What is a diaper cake?

A diaper cake is an unedible baby gift that resembles a tiered cake. It is made with rolled disposable diapers, and depending on your order, includes items such as baby blankets, hooded towels, washcloths, baby bath, lotion, powder, shampoo, pacifier, baby bottles, a onesie, nail clippers, and a toy on top.

What is a cupcake?

A cupcake is a smaller version of the diaper cake, and is more affordable. It includes 10 diapers, a pacifier, nail clippers, baby wash, baby lotion, receiving blanket, and a toy on top.

Is there a gift card included?

Yes, a personalized gift card is included with every order.

Do you have any shipping specials?

Yes, all orders $100 and over qualify for FREE shipping.

Do you ship all over the United States?

Yes, we currently ship to all physical addresses in the U.S. We are unable to ship to P.O. boxes.

Do you ship to Hospitals?

No, we do not ship to hospitals. After your order is received, the cake normally ships out within 48 hours. More than likely, the recipient of your gift will have already been released from the hospital.

How are the Diaper Cake’s shipped?

After the diaper cake is made, the tiers are carefully tied together with ribbon, and then put into a clear cellophane bag, and tied with a coordinating bow. A card with the ingredients, and a personalized gift card is included with the cake. The diaper cakes are then placed into a shipping box, and packaged for the safe arrival to the recipient.

What if I only want certain things on a cake?

We can create whatever you would like on the custom diaper cake page. Email us at info@babycakeshoppe.com with the specific items for your cake, and we will do our best to accommodate your special order. We will price out the items, and send you a email with the amount due for the cake, at that point, your order will be placed.

Are you able to work from a baby registry?

Yes, we can work from a baby registry. As long as the store is located in our area, we are able to buy the items of your choice from the registry, and make the cake from those items. All we need is a email from you, and the detailed information of the items you want included on the cake.

Baby Diaper Cupcakes

Cupcakes are a smaller version of the baby diaper cakes. They are a quality, unique baby gift that is more affordable. The cupcakes include, 10 huggies diapers, a receiving blanket, baby wash, baby lotion, a pacifier, nail clippers, and a coordinating toy as the topper. Curling ribbon completes the cupcake. The cupcake is placed into a clear cellophane bag, and tied with a bow. A personalized gift card is included, along with the ingredients of the cake.

Baby Diaper Cakes

Welcome to the Baby Cake Shoppe. We specialize in providing handmade baby diaper cakes. What is a baby diaper cake? A baby diaper cake resembles a tiered cake, made with rolled diapers, and some of the items that are necessities for the new parents. Some of those items include nail clippers, baby bath and lotion, and a baby toy is included with every cake! A complete list of ingredients is listed along with the cake of your choosing. At the Baby Cake Shoppe, all cakes are handmade at the receipt of your order. All baby items come straight from the original packaging, and are added to your cake. The only thing you have to do is complete the online order form, and we’ll take care of the rest for you.

Baby Diaper Cupcakes2 Tier Baby Diaper Cake3 Tier Baby Diaper Cake

 

Diaper cakes from the Baby Cake Shoppe are wonderful gifts for the expectant parents that adds a personal touch. Every cake is wrapped in cellophane, and a gift card is included. Diaper cakes make a wonderful centerpiece at the baby shower, and can also double as your gift for the expectant mother. Everything that comes on the cake is 100% usable for the baby. Some people prefer the cupcake size, and purchase several of them to use as centerpieces on smaller groups of tables. Baby diaper cakes are not only great baby shower gifts, but wonderful gifts that can be sent across the miles, and still have that personal touch. We offer diaper cakes for both boy’s, girl’s, and even some cakes that are neutral, and would be a great gift for either a boy or a girl. No matter when, where, or how the baby gift is given, diaper cakes make wonderful baby gifts that new parents rave about, and are very appreciative of.

Review 2, Math 2415, Ch. 12

Ace

SHORT ANSWER.  Write the word or phrase that best completes each statement or answers the question.

 

If r(t) is the position vector of a particle in the plane at time t, find the indicated vector.

1) Find the velocity vector.                                                                                                                              1)

r(t) = (cot t)i + (csc t)j

 

 

The position vector of a particle is r(t). Find the requested vector.

2) The velocity at t = π for r(t) = 5sec2(t)i - 8tan(t)j + 6t2k                                                                      2)

4

 

 

3) The acceleration at t = 3 for r(t) = (6t – 3t4)i  + (7 – t)j + (6t2 – 3t)k                                                   3)

 

 

Find the unit tangent vector of the given curve.

4) r(t) = (4 + 11t4)i + (5 + 2t4)j + (5 + 10t4)k                                                                                                   4)

 

 

Compute rʹʹ(t).

5) r(t) = (9 ln(6t))i + (8t3)j                                                                                                                                 5)

 

 

Evaluate the integral.

4

 

6)  ∫

0


-i -    2t

(5 + t2)2


j + 15

2


tk  dt                                                                                                                 6)

 

 

 

MULTIPLE CHOICE.  Choose the one alternative that best completes the statement or answers the question.

 

 

 

7)  ∫


π/4


 

(8 cos t i + 5 sin t j) dt                                                                                                                                       7)

 

-π/4

A) 8   2 i + 5   2 j                 B) 5   2 i                               C) 0                                       D) 8   2 i

 

 

Solve the problem. Assume the x -axis is horizontal, the positive y -axis is vertical (opposite g), the ground is horizontal, and only the gravitational force acts on the objects.

8) A projectile is launched from level ground at a launch angle of 26 ° and an initial speed of 48 m/sec.      8) How far away from the launch point does the projectile hit the ground?

A) ≈ 230 m                           B) ≈ 60 m                             C) ≈ 185 m                           D) ≈ 290 m

 

Horizontal range = v^2*sin2@ / g = 48^2*sin 52 / 9.8 = 2304*.788/9.8 = 185  m

 

Answer C .

 

 

Calculate the arc length of the indicated portion of the curve r(t).

9) r(t) = 4ti  +  10 cos  3 t j   +  10 sin  3 t k; 1≤ t ≤ 8                                                                                                  9)

10                        10

A) 225                                   B) 35                                     C) 45                                     D) 175

 

 

To get the arc length, we need to integrate the square root of
(dx/dt)^2 + (dy/dt)^2 + (dz/dt)^2
where r = < x(t), y(t), z(t) >

 

=  (4)^2 + ( 3 sin 10 t / 3)^2  + (-3 cos 10t /3 )^2

 

=  16  + 9  +  9  +  (sin^2 10 t / 3 + cos^210 t / 3)

 

= 16 + 18 + 1 = 35.

 

 

 

 

 

Find the length of the indicated portion of the trajectory.

10) r(t) = (etcos t)i + (etsin t)j + 3etk,  -ln 2 ≤ t ≤ 0


 

10)

 

A)     11

2

WE  DO dr/dt

 

(Sint *e^t +cos t * e^t)I  + (e^ t  – cos t  + sin t *e^ t )*j +  3 e^t * k

 

Length  would be  =  sqrt [  e^2t (sint+cost)^2 + [e^2*t(sint  - cos t ]  +  9 *e^2t .]

 

=  e ^2 t   {  2 *(sin^2t +cos^2t )  +    9  }  =  11 e^2t

 

=  [-ln2,0]  ∫sqrt 11 *e^t

 

=  sqrt11/2.


B)     7

2


C)     7

4


D)     11

4

 

 

 

 

 

 

 

 

 

 

 

 

 

Find the curvature of the curve r(t).

11) r(t) = (7 + 6 cos 9t) i - (4 + 6 sin 9t)j + 10k


 

11)

 

A) κ = 3

2

Dr /dt  = ( 54 sin9t  ) i  + 54 cos 9t j  +

|dr/dt|=  sqrt( 54*54+54*54 )  =   1.41 *54.

Now   T’(t)‘ =  54 sin 9t +54 cos 9t  /  √2*54

=  <1/√2  (sin9t) , 1/√2   (cos 9t), 0>

T ‘(t) =<9/√2 * -cos 9t ,  9/ /√2 * sin 9t , 0>

| T ‘(t)|=  √  81/2  (cos^2 9t+  sin ^2 t) =   9 / √2

K  = | T ‘(t)| / r’(t)|| 9/√2  /   √2 *54 = 1/6

Answer = D

 B) κ = 6                                C) κ =  1

36


D) κ = 1

6

 

 

 

 

 

 

 

 

Find the curvature of the space curve.

12) r(t) = (t + 8)i + 2j + (ln(sec t) + 1)k

A) κ = sin t                           B) κ = cos t                          C) κ = sec t                          D) κ = csc t

 

 

r(t) = <t+8, 2, ln(sect )+ 1>

 

r’(t) = <1,0, tant>

 

|r’(t)|=√1 + tan ^2t   =  sec^2 t  =  sec t .

 

 

T’(t)=.  <cost ,0 , sin t>

 

| T’(t)|= √ cos^2t+sin^2t  =  1

 

So  k =  1  /  sec t   = cos t

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 


 

12)

 

 

 

 

Find the unit tangent vector T and the principal unit normal vector N.

13) r(t) = (ln(cos t) + 3)i   + 10j  + (10 + t )k, -π/2 < t < π/2

A) T = (-sin t)i  + (cos t)k; N = (-cos t)i  + (sin t)k

B) T = (-sin t)i  + (cos t)k; N = (-cos t)i  - (sin t)k

C) T = (sin t)i  + (cos t)k; N = (cos t)i  - (sin t)k

D) T = (sin t)i  - (cos t)k; N = (cos t)i  - (sin t)k;

 

 

 

 

 

dr/dt  = < -tant ,  0, 1>

 

d2r/dt2 = <  sec^2t ,  0 , 0 >

 

 

r’ =  < -tant ,  0, 1>

 

|r’|=  sqrt (  1+ tan ^2t) =  sec t

 

Tangent =  r ‘ | |r ‘ |

 

T  =   sec t  < -tant , 0 , 1>

 

=   <- sin t , 0 , cos t >

 

 

T ‘  = < -cos t , 0 , -sint >

 

|T| = sqrt ( sin^2 t+ cos ^2 t)  = 1

 

N  =  1/1< -cos t  , o , -sin t>

 

 

 

Answer  is    B

 

 

 

 

 

 

 

 

 

 

 

 

 


 

13)

 

 

 

 

FInd the tangential and normal components of the acceleration.

14) r(t) = (t + 1)i + (ln(cos t) – 2)j + 3k

A) a = (-sec t tan t)T - (sec t)N                                       B) a = (sec t tan t)T - (sec t)N

C) a = (sec t tan t)T + (sec t)N                                        D) a = (-sec t tan t)T + (sec t)N

 

 

 

 

 

 

 

 

 

Dr /dt  =  i  +   -tan t  * j   +   k

 

D2r/dt2= < 0, -sec^2t ,0>

 

|r’(t)| =   1 ^2  +  tan^2  =  sqrt(  sec^2 t )  =sec t

 

r’(t) X r”(t)  =  < 0 ,  (tant)/(cos^(2)t), 0>

 

 

A Tangential=   < 0, tant *sec t,0>

 

A normal = 1/ sec t  <  sec ^2 t >    =   sec T  N

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

So ,

 

C  =

a = (sec t tan t)T + (sec t)N       is the  answer.

 

 

 


 

 

14)

 

 

 

SHORT ANSWER.  Write the word or phrase that best completes each statement or answers the question.

 

Solve the problem.

15) Find the unit tangent and normal vectors for the curve y = t2, x = t at the point t = 1.

 

<x(t),y(t)> = <t,t^2,0>

 

 

R’(t) =<1, 2t,0 >

 

|R’(t)|= √(1 + 4t^2)

 

 

 

 

 

T(t) =  unit tangent vector =  1/√(1 + 4t^2) i + 2 *t//√(1 + 4t^2)

 

at  t  = 1

=  √5/ 5   i+ 2√5 / 5   j

 

 

Now ,

 

T’(t) =  -4 t/(4t^2+1)^3/2 i   +  2 //(4t^2+1)^3/2 j

 

Now  |T(t)| =  √  16 / (4t^2+1) ^3 +  4/(4  t^2+1)^3

 

Now    N(t)  = 4 t/(4t^2+1)^3/2 i   +  2 //(4t^2+1)^3/2 j   /   √  16 / (4t^2+1) ^3 +  4/(4  t^2+1)^3

 

At  t  =  1

 

=   -4 /(5)^3/2 i +  2 / 5^3/2  j /  √[1/ (5^3) {  20 }]

 

=   1/5 * 2 √5 i -  √5/5j

 

 

 

 

 

16) Find the curvature of x = 2t -3, y = t2 + 2t + 5 at the point where t = 0.

 

Curvature   dy/dx  =    2  /  2 *t  + 2  + 5  =  2 /2  = 1

 

 

 

 

16)

 

 

 

17)


 

 

Evaluate the integral  ∫

0


 

π/2


 

 

tsin t2 i dt .

 

 

=  -1/2  * cos  t^2

 

=

 

 

 


17)

 

 

 

 

18) A particle moves in space. Its position vector at time t is r(t) = ti + j sin t + k cos t. Find the velocity and acceleration vectors at time t = 0.

 

Dr/dt  =  1* i  +  -cos t * j  +  k  * sin t .

 

So   velocity   at  t = 0

=   <1 , -1  , 0>

 

Acceleration  =   < 0 ,  sin t  ,  -cos t >  =  < 0,0,-1>

 


18)

 

 

 

MULTIPLE CHOICE.  Choose the one alternative that best completes the statement or answers the question.

 

Solve the problem. Unless stated otherwise, assume that the projectile flight is ideal, that the launch angle is measured from the horizontal, and that the projectile is launched from the origin over a horizontal surface

 

19) A projectile is fired at a speed of 920 m/sec at an angle of 30°.  How long will it take to get 23 km downrange?  Round your answer to the nearest whole number.

A) 27 sec                                                                              B) 31 sec

C) 29 sec                                                                             D) It will never get that far downrange.

 

 

Horizontal speed

 

V cos 30  m/sec.

 

For  23 km  it will take = 23 *1000 / 920 * cos 30 =  29 sec


19)

 

 

 

 

20) A spring gun at ground level fires a tennis ball at an angle of 35°. The ball lands 24 m away. What was the ballʹs initial speed? Round your answer to the nearest tenth.

A) 15.8 m/sec                       B) 5.1 m/sec                         C) 250.3 m/sec                    D) 20.2 m/sec

 

 

24 = v^2 * sin 7 0 / 9.8

 

So  v^2  =  24 *9.8 / sin 70 = 15.8 m/sec
20)

 

Answer Key

Testname: REV2M2415

 

 

 

1) v = (-csc2 t)i - (cot t csc t)j

 

2) v π

4


 

= 20i - 16j + 3πk

 

3) a(3) = -324i + 12k

4) T = 11 i +  2 j + 2 k

15      15      3

5) rʹʹ(t) = – 9 i + 48tj

t2

6) -4i - 16 j + 40k

105

7) D

8) C

9) B

10) A

11) D

12) B

13) B

14) C

15) T =     5 i + 2   5 j;      N = 2   5 i -   5 j

5           5                        5           5

16)     1   

4   2

17)   2 –  2  j

4

18) v(0) = <0, 1, 0>

a(0) = <0, 0, -1>

19) C

20) A

 

http://en.wikipedia.org/wiki/Calculus

MATH 108 Fall 2013 HW 2

DesertSimplify the expression completely.

 

1)

=21*cos^(3)x*(sinx)/(7*sinx*sinx*cosx)

 

21*cos^(3)x*(sinx)/(7sinx*sinx*cosx)

 

21*cos^(3)x*(sinx)/(7sin^(2)x*cosx)

 

21*cos^(3)x*(sinx)/(7cosxsin^(2)x)

 

21*cos^(3)x*(1)/(7sinxcosx)

 

21*cos^(3)x*(1)/(7cosxsinx)

 

21*cos^(2)x*(1)/(7sinx)

 

21cos^(2)x*(1)/(7sinx)

 

(21cos^(2)x)/(7sinx)

 

(3cos^(2)x)/(sinx)

 

3cosxcotx

 

2)        = 1/ (cosx+sinx) * [4/(sinx+cosx)+7]= 1/ (cosx+sinx) * [4+sinx+cosx]  =  1/(cos^2x – sin^2x) (4 + 7 sinx + 7cosx)

 

3)  = 10 / 5  *  tan^3x/tanx *  sec^2x / sec x = 2 * tan^2x *secx

 

4) =  (cos^2x – 2 cosx  – 8 ) / (cosx -4) = cos^2x – 2cosx +1  – 9  = (cosx-1)^2  – 3 ^2  /(cosx-4) =( cos x +2)

 

 

Find the exact circular function value.

5) cos 2π/3 = cos (Π -Π/3) = -cos 60 = -.5

 

6) tan 5π/6 = tan (Π –Π/6) = – cot 30 = -.577.

 

Graph.

7) y = sin(x + 3π/2)

 

 

Find the amplitude, period or phase shift.

8) Find the period and phase shift of y =

y=Acos(Bx-C), A=amplitude, Period=2π/B, Phase Shift=C/B

period = 2 * π / ¼ = 8 / π.

 

9) Find the amplitude of y = 5 cos (x – π).

Amplitude a = 5.

y=Asin(Bx-C), A=amplitude, Period=2π/B, Phase Shift=C/B

 

10) Find the period of y = 2 cos (2x + π/3).

Period = 2 * π / 2 = π.
11) Find the phase shift of y = -4 + 3sin (5x – π/6)

Phase shift =  π/6 /5  = π/30.

 

Find the trigonometric function value of angle θ.

12)  sin θ = – 5/13 and θ in quadrant III

Find sec θ and csc θ

 

Cos @ = sqrt (1 – (5/13)^2) = 12/13

Sec @ = 13/12

Cosec @= -13/5.

 

13) cot θ = – ¾ and θ in quadrant II

Find csc θ and sinθ

 

Cosec^2 @ = 1 + 9/16 = 25/16

Cosec @ = 5/4

Sin @=  4/5.

 

 

15) Find  exactly in degrees. = 210 degree

 

16) Find  exactly in radians. = Π/3.

 

Check out http://www.acemyhw.com

MAT 111

   MAT 111-03                                     Test 1(take-home)                    Handed out Friday, September 20          

Due  by  Thursday, September 26, noon (12:00 pm)

Late papers will be penalized:

–10% if I receive your paper after 12:00pm but before 4:00pm Thursday, September 26

–25% if I receive your paper after 4:00pm Thursday, but before 9:00am Friday, September 27

–50% if I receive your paper after 9:00am Friday, but before 3:00pm Friday, September 27

–100% if I receive your paper after 3:00pm Friday, September 27

************************************************************************************

Instructions

  • Enter only the neat, final version of your work, together with your answers, on the spaces provided

on these sheets. ALL OF YOUR WORK AND YOUR ANSWERS SHOULD BE ON THESE

SHEETS I AM PROVIDING.

 

  • Enter your answers in the boxes provided.

 

  • You must show supporting work for full credit.

 

  • Although you may consult others while working on these problems, the final work you submit

must be written by you.

 

 

 

 

 

 

 

 

 

By signing below, I am acknowledging that I have read the above Instructions, and agree to abide by them. I understand that failure to follow any of these Instructions may result in a lowering of my test score.

 

Your name:

 

 

Your Signature: _____________________________

 

 

 

 

 

 

 

 

 

Let   A, B, C, and D be the last four digits of your 8-digit student number, written in decreasing order (from the largest to the smallest).  (So if  your student number is  111-23-546, then for your test

 A = 6, B = 5, C = 4 and D = 3. If your student number is  101-24-003, then  A= 4, B=3, C=0, and D=0.)

101-43-030

 

Write your numbers here:   A   =  __4__   ,     B =  __3__  ,       C =  __1___,      D =  __0___

 

Warning. You must use your student number. 

 

 

1.   A store has shoes that have their sale price  marked down by 20%.  A week later their sale price is reduced by

10%.  The following week the shoes have their sale price reduced by 50%.

 

If the final price (before tax) of the shoes is  $AB.CD (that is, Ax$10 + Bx$1 +  Cx10¢ + Dx1¢), then

what was their price before the first discount?

 

Let  the price of  shoes be $x

Now sale price = .80 *x .

After one week its price = .80*x * .90 = .72 x.

Now the following week  price = .72 * 50% = .36 *x .

 

Now  the

Final price = $AB.CD

So value of x =  100/36 * $AB.CD =  $ 25/9 *AB.CD

 

 

For problems 2. and 3.,    r %=  A.BCD %  for your A, B, C, and D . (So, for example,  if A = 6, B = 3,

C = 3,  and  D = 1, then   r % = 6.331%).

 

 

  2.  (a)  If  $1500 is invested at an annual simple interest rate of  r %,then  how much will it grow to

in 64 months?  Write your answer in dollars+cents, rounded  to the nearest cent.

 

SI =  P * R /100 * T

 

64 months = 5 years 4 month = 16/3 year

 

So,  interest  =  16/3*1500*r/100 =  $ 80 *r .

 

For example  r= 5.431 %

So  interest =  $434 and 43 cents

 

 

 

$434 and 43 cents

 

 

 

(b)  If  $1500 is invested at an annual interest rate of   r %  compounded monthly, how much will it

grow to in 64 months?  Write your answer in dollars+cents, rounded  to the nearest cent.

 

 

 

Amount = P * (1+ r/12)^12

= 1500 * (1+ 5.431/12)^12

= $ 2002.64

 

 

So, compound  interest  = 2002.64- 1500 = $502 and 64 cents

 

 

 

 

 

 

 

                                                                                                                                    $502 and 64 cents

 

 

 

 

 

 

 

3.  How many pennies would you have if together they weigh  r%  of a ton?  Round your answer to the nearest whole

number of pennies.

 

a ton = 1000 Kg

let   r = 6.250 %

 

so , I  would  have  = 6.250 /100 * 100 kg

 

assuming each  penny  is  2.5 gm

 

so  number of penny = 6.250 /100 * 100 * 1000 / 2.5  = 6.250 / 2.5 * 1000 = 2.5*1000 = 25000  pennies

 

 

 

 

 

25000  pennies

 

 

 

 

 

 

 

 

 

 

 

 

 

            4.  There are three faucets available to fill a wading pool. The first alone would take 90 minutes to fill

the pool. The second alone would take 75 minutes to fill the pool. The third alone would take 1 hour

to fill the pool. How long would it take to fill the pool if all three faucets are used at the same time?

Write your answer in minutes+seconds, rounded to the nearest second

 

 

Capacity  of  1st faucet = 1/90 .

 

Capacity of2nd faucet = 1/75.

 

Capacity of  3rd faucet = 1/60.

 

If   they are used at same time so total capacity = 1/60 +1/75 + 1/90 = 37/900

 

 

So,  the  time taken to fill the pool = 1 / (37/900)= 900/37 = 24 minutes and 19 seconds

.

 

 

 

 

 

 

 

24 minutes and 19 seconds

 

5.  A chessboard, like a checkerboard, has 64 squares, usually 32 white and 32 black. Imagine putting a penny

on the 1st white square, two pennies on the 2nd white square, four on the 3rd white square, and so on for all

of the 32 white squares. Thus each white square but the 1st contains twice as many pennies as the previous

white square. Assuming each penny weighs 2.5 grams, then how heavy a stack could be made from all

of the pennies? Round your answer to the nearest ton. Write your answer in tons+pounds, rounded to the

nearest pound.

 

 

 

 

 

1st  white  square  has  one penny

2 nd  white square  has  2

3rd  white  square has 4

 

So ,    n  white square has 2^(n-1).pennies

 

So  for  32  squares  the  number  of  pennies  will be  = ∑ 2^(n-1)  for n = 1 to 32.

 

=  4294967295  pennies  can be kept .

 

Now  ,

 

Total  weight  = 4294967295 *2.5 gm

 

Now  1 ton = 1000 * 1000  gm

 

So  weight  = 4294967295 *2.5 /  10^6=  10737.42  ton.

= 10737  and 840 pounds.

 

 

 

 

 

 

 

10737  and 840 pounds

 

 

 

 

 

For Problem 6, use  I %  = r % + 10 %  as your annual interest rate, where  r % = A.BCD % 

         is the same interest rate you used in Problems 1 and 2.

 

 

6.  The following table shows all transactions on a particular credit card during the month of

     November.

Assume an annual interest rate of   I %, and assume a balance of $1916.02  at the start of

November. (Thus a payment of $1916.02  at the start of November would reduce the debt to $0.)

 

      Date                             Transaction
   Nov   3                $2.07     video rental charged to card
   Nov  10                $56.18   payment for cell phone bill paid for with the credit card
   Nov  12                $300      payment to credit card
   Nov  21                $35.59   pants which were charged to card in August are returned to

Walmart, and the money is credited to the card

   Nov  30               $ 74.26   groceries charged to card

Using the average daily balance method, calculate the interest (finance charge)  for November, and

the balance due at the start of December.   Write your answer in dollars+cents, rounded to the nearest cent.

 

I % = 10 + 5.431 = 15.431 %

 

At   start  of November  =  $1902.

 

Amount  paid with credit  card = 2.07  +  56.18 + 74.26 = $ 132.51

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

                                                                      November interest

 

 

 

                                       Balance due at the start of December