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2 years ago

Work out the answer to this problem to reduce to the simplest form 5/8-1/2=

Mathematics

2 answers:

Mamont248 [21]2 years ago

8 0

The answer in simplest form is 1/8

inysia [295]2 years ago

3 0

In order to subtract fractions, you need the denominators (bottom number) to be the same. $\frac{1}{2} * \frac{4}{4} = \frac{4}{8}$ . $\frac{5}{8} - \frac{4}{8} = \frac{1}{8}$

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0.66x10 to the 3rd power

Bas_tet [7]

The answer to this is 660

4 0

2 years ago

Read 2 more answers

Find the value of the expression 4a4 − 2b2 + 40 when a = 2 and b = 7.

Makovka662 [10]

4 a⁴ - 2 b² + 40 :

4 * ( 2⁴) - 2 * ( 7²) + 40 =

4 * 16 - 2 * 49 + 40 =

64 - 98 + 40 =

-34 + 40 =

+ 6

<span>hope this helps!</span>

4 0

3 years ago

Mx + 4y = 3t solve for x

olasank [31]

Mx + 4y = 3t
mx = -4y + 3t
x = (-4y + 3t) / m or x = -4y/m + 3t/m

6 0

3 years ago

A rectangular field is to be enclosed by a fence and divided

Phantasy [73]

Answer:

80000 square meters

Step-by-step explanation:

perimeter + dividing fence = 800

let a = length

let b = width

let c = length of dividing fence

perimeter = 2*a + 2*b

let's say...

c is the same as the length

2a + 2b + c = 800

2a + 2b + a = 800

3a + 2b = 800

area = length*width

area = a*b

area / b = a

3*(area/b) + 2b = 800

3*(area/b) = 800 - 2b

area/b = (800 - 2b)

area = (800 - 2b)*b

To make the area large, we make the right hand side large.

800b - 2b^2

If you put in terms of x, y it looks like a downward opening parabola, so the max area is at the vertex. Half way between the roots.

y = -2x^2 + 800x

y = -x^2 + 400x

0 = x*(-x + 400)

roots are x= 0 and x = 400

vertex is at x, aka b = 200

area at b=200 is (800 - 400)*200 = 80000

and a is area/b... 80000/400 = 200

5 0

2 years ago

Find the critical points of the function f(x, y) = 8y2x − 8yx2 + 9xy. Determine whether they are local minima, local maxima, or

NARA [144]

Answer:

Saddle point: $(0,0)$

Local minimum: $(\frac{3}{8}, -\frac{3}{8})$

Local maxima: $(0,-\frac{9}{8})$ , $(\frac{9}{8},0)$

Step-by-step explanation:

The function is:

$f(x,y) = 8\cdot y^{2}\cdot x -8\cdot y\cdot x^{2} + 9\cdot x \cdot y$

The partial derivatives of the function are included below:

$\frac{\partial f}{\partial x} = 8\cdot y^{2}-16\cdot y\cdot x+9\cdot y$

$\frac{\partial f}{\partial x} = y \cdot (8\cdot y -16\cdot x + 9)$

$\frac{\partial f}{\partial y} = 16\cdot y \cdot x - 8 \cdot x^{2} + 9\cdot x$

$\frac{\partial f}{\partial y} = x \cdot (16\cdot y - 8\cdot x + 9)$

Local minima, local maxima and saddle points are determined by equalizing both partial derivatives to zero.

$y \cdot (8\cdot y -16\cdot x + 9) = 0$

$x \cdot (16\cdot y - 8\cdot x + 9) = 0$

It is quite evident that one point is (0,0). Another point is found by solving the following system of linear equations:

$\left \{ {{-16\cdot x + 8\cdot y=-9} \atop {-8\cdot x + 16\cdot y=-9}} \right.$

The solution of the system is (3/8, -3/8).

Let assume that y = 0, the nonlinear system is reduced to a sole expression:

$x\cdot (-8\cdot x + 9) = 0$

Another solution is (9/8,0).

Now, let consider that x = 0, the nonlinear system is now reduced to this:

$y\cdot (8\cdot y+9) = 0$

Another solution is (0, -9/8).

The next step is to determine whether point is a local maximum, a local minimum or a saddle point. The second derivative test:

$H = \frac{\partial^{2} f}{\partial x^{2}} \cdot \frac{\partial^{2} f}{\partial y^{2}} - \frac{\partial^{2} f}{\partial x \partial y}$

The second derivatives of the function are:

$\frac{\partial^{2} f}{\partial x^{2}} = 0$

$\frac{\partial^{2} f}{\partial y^{2}} = 0$

$\frac{\partial^{2} f}{\partial x \partial y} = 16\cdot y -16\cdot x + 9$

Then, the expression is simplified to this and each point is tested:

$H = -16\cdot y +16\cdot x -9$

S1: (0,0)

$H = -9$ (Saddle Point)

S2: (3/8,-3/8)

$H = 3$ (Local maximum or minimum)

S3: (9/8, 0)

$H = 9$ (Local maximum or minimum)

S4: (0, - 9/8)

$H = 9$ (Local maximum or minimum)

Unfortunately, the second derivative test associated with the function does offer an effective method to distinguish between local maximum and local minimums. A more direct approach is used to make a fair classification:

S2: (3/8,-3/8)

$f(\frac{3}{8} ,-\frac{3}{8} ) = - \frac{27}{64}$ (Local minimum)

S3: (9/8, 0)

$f(\frac{9}{8},0) = 0$ (Local maximum)

S4: (0, - 9/8)

$f(0,-\frac{9}{8} ) = 0$ (Local maximum)

Saddle point: $(0,0)$

Local minimum: $(\frac{3}{8}, -\frac{3}{8})$

Local maxima: $(0,-\frac{9}{8})$ , $(\frac{9}{8},0)$

4 0

3 years ago