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2 years ago

Can you help me i dont know thw anwser

Mathematics

2 answers:

Margaret [11]2 years ago

8 0

$\text{Let the side of square 1, square 2 and square 3 be x, y and z respectively.}\\\\\text{Given that,}\\\\\text{Perimeter of square 1,~~ p =100 units.}\\\\\text{So}~ x=\dfrac p 4 = \dfrac{100}4 =25~\text{units.}\\\\\\\text{Area of square 2, a = 225 units}^2\\\\\text{So}~y = \sqrt{a} = \sqrt {225} = 15~ \text{units.}\\\\\text{Use Pythagorean theorem to find z}^2\\\\x^2 =y^2 +z^2 \\\\\implies 25^2 = 15^2 +z^2\\\\\implies z^2 = 625 - 225 = 400\\\\\\$

$\text{Hence the area of square 3 is 400 units}^2.$

Verdich [7]2 years ago

6 0

Answer:

400 square units

Step-by-step explanation:

The side length of square 1 is 1/4 of its perimeter, so is 100/4 = 25 units. The area of square 1 is the square of the side length, so is 25² = 625 square units.

The area of square 3 is the difference in the areas of squares 1 and 2. This is due to the Pythagorean theorem.

area 3 = area 1 - area 2

area 3 = 625 -225

area 3 = 400 . . . . square units

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Rewraite 3/7 whit a denominator of 28

deff fn [24]

Answer:

12/28

Step-by-step explanation:

3 x 4 = 12

7 x 4 = 28

4 0

2 years ago

Read 2 more answers

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)