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Svet_ta [14]
2 years ago
9

What is the present value of $6,000 to be received at the end of 6 years if the discount is 12%

Mathematics
2 answers:
ycow [4]2 years ago
8 0
Future value = FV = $6000

Time period = n = 8

interest rate = r = 12%

Present value (PV) is calculated using the formula:

PV = FV/(1+r)n

Present Value = PV = 6000/(1+12%)8 = 6000/1.128 = 2423.299368

Answer -> 2,423

Marizza181 [45]2 years ago
4 0

Answer:

$3,039.79

Step-by-step explanation:

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A pyramid of logs has 2 logs in the top row, 4 logs in the second row from the top, 6 logs in the third row from the top, and so
mamaluj [8]

Answer:

1) Is the pattern an arithmetic sequence?​

Yes it is

2)Identify a and d.​

a = First term = 2

d = Common difference = 2

3) Write the 50th term of the sequence.​

50th term = 100

4) Find the total number of logs in the first 10 rows.​

= 1010 logs

Step-by-step explanation:

Is the pattern an arithmetic sequence?​

Yes it is

2) Identify a and d.​

A pyramid of logs has 2 logs in the top row, 4 logs in the second row from the top, 6 logs in the third row from the top, and so on,

The formula for arithmetic sequence =

an = a+ (n - 1)d

a = First term

d = Common difference

For the above question:

a = 2

d = Second term - First term

= 4 - 2

d = 2

3) Write the 50th term of the sequence.​

Using the formula for arithmetic sequence

an = a+ (n - 1)d

a = 2

n = 50

d = 2

a50 = 2 + (50 - 1)2

= 2 + (49)2

= 2 + 98

= 100

The 50th term = 100

4)Find the total number of logs in the first 10 rows.​

Sum of first n terms = n/2(a + l)

n = 10

a = first term = 2

We are told that there are 200 logs in the bottom row, hence:

l = last term = 200 logs

Hence,

Sn = 10/2×[ (2 + 200

= 5(202)

= 1010 logs

7 0
3 years ago
PLEASE HELP ASAP 15 POINTS
stepan [7]

Answer: Its C

Step-by-step explanation: Hope this Helped!!!

5 0
3 years ago
Vanessa draws one side of equilateral ΔABC on the coordinate plane at points A(–2, 1) and B(4,1). What are the possible coordina
grandymaker [24]

Important question. No menu of choices given so we'll just have to figure it out.

It's not possible in two dimensions for all three vertices of an equilateral triangle to be lattice points, i.e. points with integer coordinates.  

Since A(-2,1) and B(4,1) have integer coordinates, our other point won't. There will be a √3 involved, the algebraic tell that there's a 30/60/90 triangle lurking somewhere in the problem.

Here the given side is parallel to the x axis which makes things easier.  

|AB| = 4 - -2 = 6

The x coordinate of the third vertex will be the same as that of the midpoint of AB, let's call it M,

M = ( (-2 + 4) / 2, (1 + 1)/2 ) = (1, 1)

We're making some progress.  Whatever our height h is our two candidates for the third vertex C are (1, 1 ± h).

Since |AB|=6, we get |AB|=|BC|=|AC|=6 because it's an equilateral triangle.

The altitude CM bisects the triangle into 2 30/60/90 triangles.  Let's take one of them, AMC.  Angle AMC is a right angle, so we have a right triangle with legs |AM|=3 and |CM|=h, and hypotenuse |AC|=6. The Pythagorean Theorem tells us:

3² + h² = 6²

9 + h² = 36

h² = 27

h = 3√3

Answer: C is (1, 1 + 3√3) or (1, 1 - 3√3)

4 0
3 years ago
What is the relationship between 0.04 and 0.004
Molodets [167]
0.04 is multiplied 10 times to get 0.004
4 0
3 years ago
Complete the assignment on a separate sheet of paper<br><br> Please attach pictures of your work.
Irina18 [472]

Answer:

<u>TO FIND :-</u>

  • Length of all missing sides.

<u>FORMULAES TO KNOW BEFORE SOLVING :-</u>

  • \sin \theta = \frac{Side \: opposite \: to \: \theta}{Hypotenuse}
  • \cos \theta = \frac{Side \: adjacent \: to \: \theta}{Hypotenuse}
  • \tan \theta = \frac{Side \: opposite \: to \: \theta}{Side \: adjacent \: to \: \theta}

<u>SOLUTION :-</u>

1) θ = 16°

Length of side opposite to θ = 7

Hypotenuse = x

=> \sin 16 = \frac{7}{x}

=> \frac{7}{x} = 0.27563......

=> x = \frac{7}{0.27563....} = 25.39568..... ≈ 25.3

2) θ = 29°

Length of side opposite to θ = 6

Hypotenuse = x

=> \sin 29 = \frac{6}{x}

=> \frac{6}{x} = 0.48480......

=> x = \frac{6}{0.48480....} = 12.37599..... ≈ 12.3

3) θ = 30°

Length of side opposite to θ = x

Hypotenuse = 11

=> \sin 30 = \frac{x}{11}

=> \frac{x}{11} = 0.5

=> x = 0.5 \times 11 = 5.5

4) θ = 43°

Length of side adjacent to θ = x

Hypotenuse = 12

=> \cos 43 = \frac{x}{12}

=> \frac{x}{12} = 0.73135......

=> x = 12 \times 0.73135.... = 8.77624.... ≈ 8.8

5) θ = 55°

Length of side adjacent to θ = x

Hypotenuse = 6

=> \cos 55 = \frac{x}{6}

=> \frac{x}{6} = 0.57357......

=> x = 6 \times 0.57357.... = 3.44145.... ≈ 3.4

6) θ = 73°

Length of side adjacent to θ = 8

Hypotenuse = x

=> \cos 73 = \frac{8}{x}

=> \frac{8}{x} = 0.29237......

=> x = \frac{8}{0.29237.....} = 27.36242..... ≈ 27.3

7) θ = 69°

Length of side opposite to θ = 12

Length of side adjacent to θ = x

=> \tan 69 = \frac{12}{x}

=> \frac{12}{x} = 2.60508......

=> x = \frac{12}{2.60508....}  = 4.60636.... ≈ 4.6

8) θ = 20°

Length of side opposite to θ = 11

Length of side adjacent to θ = x

=> \tan 20 = \frac{11}{x}

=> \frac{11}{x} = 0.36397......

=> x = \frac{11}{0.36397....}  =30.22225.... ≈ 30.2

5 0
2 years ago
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