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3 years ago

What is the present value of an asset that yields $1,500 in five years and $3,000 in 10 years if the discount rate is 8%

Mathematics

1 answer:

Eddi Din [679]3 years ago

7 0

I believe the present value would be $1,389.58

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Find the 6th term in the sequence

Dimas [21]

Answer:

C.64

Step-by-step explanation:

The first step is to figure out what the sequence is, the first step is 2, 4 the only ways for 2 to get to 4 would be +2 or *2 so we will look at the next step, 4 to 8. The only ways for 4 to get to 8 is +4 or *2, since these both have *2 in common we will check that with all of the terms

2, 4, 8, 16, 32

2 (*2) = 4 (*2) = 8 (*2) = 16 (*2) 32

Since the equation is working we are going to multiply 32 by 2 to get the 6th term

32 (*2) = 64

6 0

2 years ago

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the area of a rectangle of a barn is 200 square feet. the length is 10 feet longer than width . find the length and width of the

Elina [12.6K]

Area of rectangular barn = 200 (sqft)

Length= Width + 10

Length x Width = 200 ==> (Width+10)Width = 200

Width² +10Width -200 = 0===> width = 10 ===> length = 20

8 0

3 years ago

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I have a 84% in class, there is a 25 point quiz worth 100 points, how what grade do I need to get to at least have an average gr

Thepotemich [5.8K]

At least about 20 points it can depend on the class but not really.

4 0

3 years ago

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If M=-7 and B=9 what are the answers to the following 67xM= -23xB

alekssr [168]

Answer: -207

Step-by-step explanation:

8 0

2 years ago

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The triangle T has vertices at (-2, 1), (2, 1) and (0,-1). (It might be an idea to

Firdavs [7]

Rewrite the boundary lines y = -1 - x and y = x - 1 as functions of y :

y = -1 - x ==> x = -1 - y

y = x - 1 ==> x = 1 + y

So if we let x range between these two lines, we need to let y vary between the point where these lines intersect, and the line y = 1.

This means the area is given by the integral,

$\displaystyle\iint_T\mathrm dA=\int_{-1}^1\int_{-1-y}^{1+y}\mathrm dx\,\mathrm dy$

The integral with respect to x is trivial:

$\displaystyle\int_{-1}^1\int_{-1-y}^{1+y}\mathrm dx\,\mathrm dy=\int_{-1}^1x\bigg|_{-1-y}^{1+y}\,\mathrm dy=\int_{-1}^1(1+y)-(-1-y)\,\mathrm dy=2\int_{-1}^1(1+y)\,\mathrm dy$

For the remaining integral, integrate term-by-term to get

$\displaystyle2\int_{-1}^1(1+y)\,\mathrm dy=2\left(y+\frac{y^2}2\right)\bigg|_{-1}^1=2\left(1+\frac12\right)-2\left(-1+\frac12\right)=\boxed{4}$

Alternatively, the triangle can be said to have a base of length 4 (the distance from (-2, 1) to (2, 1)) and a height of length 2 (the distance from the line y = 1 and (0, -1)), so its area is 1/2*4*2 = 4.

6 0

2 years ago