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4 years ago

How long does it take to go 5.067km at 225mph?

1 answer:

3 0

It would take .0139 hours or.834 mins to travel 5.067Km at 225 mph.

First you have to convert kilometers to miles which is 1.6 kilometer are in 1 mile then multiply 225 times 1.6 which gives you 362.25KPH then divide 5.067Km by 362.25

Hope this helps

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A women has 157.5 different ribbons. How many different ribbons does she have?

kykrilka [37]

Answer:

158 ribbons even half of one still counts.

Step-by-step explanation:

7 0

3 years ago

Find two numbers x and y such that a) 2x+y=100 and A=2x+2xy+y is maximized b) 2x+4y-15=0 and B= √x2+y2is minimized. Note that in

zaharov [31]

Answer:

a) x = 25, y = 50

b) x = 1.5, y = 3

Step-by-step explanation:

We have to use Lagrange Multipliers to solve this problem. The maximum of a differentiable function f with the constraint g(x,y) = b, then we have that there exists a constant $\lambda$ such that

$\nabla f(x,y) = \lambda \, \nabla g(x,y)$

Or, in other words,

$f_x(x,y) = \lambda \, g_x(x,y) \\ f_y(x,y) = \lambda \, g_y(x,y)$

a) Lets compute the partial derivates of f(x,y) = 2x+2xy+y. Recall that, for example, the partial derivate of f respect to the variable x is obtained from derivating f thinking the variable y as a constant.

$f_x(x,y) = 2 + 2y$

On the other hand,

$f_y(x,y) = 2x+1$

The restriction is g(x,y) = 100, with g(x,y) = 2x+y. The partial derivates of g are

$g_x(x,y) = 2; g_y(x,y) = 1$

This means that the Lagrange equations are

$2y + 2 = 2 \, \lambda$
$2x +1 = \lambda$
$2x + y = 100$ (this is the restriction, in other words, g(x,y) = 100)

Note that 2y + 2, which is $2 \, \lambda$ is the double of 2x+1, which is $\lambda$ . Therefore, we can forget $\lambda$ for now and focus on x and y with this relation:

2y+2 = 2 (2x+1) = 4x+2

2y = 4x

y = 2x

If y is equal to 2x, then

g(x,y) = 2x+y = 2x+2x = 4x

Since g(x,y) = 100, we have that

4x = 100

x = 100/4 = 25

And, therefore y = 25*2 = 50

Therefore, x = 25, Y = 50.

b) We will use the suggestion and find the minumum of f(x,y) = B² = x²+y², under the constraing g(x,y) = 0, with g(x,y) = 2x+4y-15. The suggestion is based on the fact that B is positive fon any x and y; and if 2 numbers a, b are positive, and a < b, then a² < b². In other words, if (x,y) is the minimum of B, then (x,y) is also the minimum of B² = f.

Lets apply Lagrange multipliers again. First, we need to compute the partial derivates of f:

$f_x(x,y) = 2x \\f_y(x,y) = 2y$

And now, the partial derivates of g:

$g_x(x,y) = 2 \\ g_y(x,y) = 4$

This gives us the following equations:

$2x = 2 \, \lambda \\ 2y = 4 \, \lambda \\ 2x+4y-15 = 0$

If we compare 2x with 2y, we will find that 2y is the double of 2x, because 2y is equal to $4 \, \lambda$ , while on the other hand, $2x = 2 \, \lambda$ . As a consequence, we have