11.

At the start of 2015, barry’s house was worth £270 000

Likurg_2 [28]

Answer:

313003.6

Step-by-step explanation:

The number increases by 3 percent every year so 313003.6 is it.

3 0

2 years ago

Two parallel lines are cut by a transversal,

Volgvan

Answer:

also 75°

Step-by-step explanation:

the angles across each other at a crossing point of 2 lines must be equal.

so, 2 = 3.

and the angles at the crossing point with a parallel line are simply a "mirror" or rather a copy of the angles at the crossing point with the first line.

so, 2 = 3 = 6 = 7

and 1 = 4 = 5 = 8

7 0

2 years ago

I need help for the solution

Sonja [21]

Answer:

$\boxed{ \ dY_t=(2\theta+2\psi Y_t+\phi^2)dt+2\phi \sqrt{Y_t}dW_t\ }$

Step-by-step explanation:

it is a long time I have not applied Ito's lemma

I would say the following

for $f(x)=x^2$

f'(x)=2x

f''(x)=2

so using Ito's lemma we can write that

$dY_t=2V_tdV_t+\phi^2dt$

$dY_t=2(\theta+\psi V_t^2)dt+2\phi V_tdW_t+\phi^2dt$

$dY_t=(2\theta+2\psi V_t^2+\phi^2)dt+2\phi V_tdW_t$

so it comes

$dY_t=(2\theta+2\psi Y_t+\phi^2)dt+2\phi \sqrt{Y_t}dW_t$

3 0

3 years ago

Calculate the sum of 1/4 and 1/7

KonstantinChe [14]

Try to find the lowest common denominator. In this case it's 28, which makes our numbers as follows:

1/4 = 7/28
1/7 = 4/28

7/28 + 4/28 = 11/28

We can't simplify it so that's the result.

7 0

3 years ago

A rectangular cow pasture is enclosed on three sides by a fence and the fourth side is part of the side of a barn that is $400$

Degger [83]

Let us list out the given information:

Cost per foot = $5

Total fencing cost = $1200

Let L be the length of the barn that will maximize the area of the pasture and W be the width of the rectangular cow pasture.

From the given information, we need to do the fencing for three sides covering 1 L and 2 W's because other L will be the fourth side is part of the side of a barn. We need to write the equation based on the cost information.

If cost/foot is $5, then cost of length ' L ' will be 5L dollars and cost of '2W' width equals 10W dollars. Total cost is $1200. We can set up second equation as

5L + 10W = 1200. This can be simplified by dividing throughout the equation by 5 to get L+ 2W= 240. Solving for L we get L = 240 - 2W

Area of a rectangle is given by A = L ×W.

Substituting 240 - 2W for L we get the area function as..

A=(240 - 2W)×W= 240W-2W²

A = -2W²+240W.

When we have a function of the form f(x) = ax^2 + bx + c, then maximum value of the function can be obtained for x = -b/2a. Using that idea we can find the maximum area.

Here in the area function a = -2 and b = 240

For W = -b/2a = -240/2(-2) = -240/-4 = 60 feet, we get maximum area.

The question is asking us to find Length. We can substitute W = 60 in the equation L = 240 - 2W to find the Length.

L = 240 - 2(60) = 240 - 120 = 120 feet.

Conclusion: The length of the side parallel to the barn that will maximize the area of the pasture is 120 feet.

Note: Here the information 400 feet in the sentence "the fourth side is part of the side of a barn that is 400 feet long" given to distract us, we might not require that information to find the Length.

5 0

3 years ago