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2 years ago

The perimeter of the polygon is 36 in. What is the length of side z? any one for 20pt

Mathematics

1 answer:

Reil [10]2 years ago

7 0

Answer:

brqnficojeuiyrjnxsoihgfyufhrbjenfrvbwejifwhbfuiwebf

Step-by-step explanation:

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Solve the word problem and show your work. I need help

Aleksandr [31]

Answer:

ima just take all your points and roll out tbh

Step-by-step explanation:

6 0

2 years ago

Do yall know any easy way to do slope

mart [117]

Answer:

To find the slope of two given points, you can use the point-slope formula of (y2 - y1) / (x2 - x1). With the points plugged in, the formula looks like (3 - 2) / (4 - 1). Simplify the formula to get a slope of ⅓.

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

6+[11-{10+2-(8 of 2+1-3×4) }]

Vesna [10]

Answer:

Step-by-step explanation:

Step-by-step explanation:

6+[11-{10+2-(8×2+1-3×4)}]

6+[11-{10+2-(8×2+1-3×4)}]6+[11-{10+2-(12)}

6+[11-{10+2-(8×2+1-3×4)}]6+[11-{10+2-(12)} ]6+[11-[10+2-12}]

6+[11-{10+2-(8×2+1-3×4)}]6+[11-{10+2-(12)} ]6+[11-{10+2-12}]6+[11-{0}]

6+[11-0]

6+[11]

6+11

17 Ans.

7 0

2 years ago

Problemas de razonamiento división de números decimales. Ayer Susana se fue de viaje a visitar a unos familiares. Recorrió 135,7

schepotkina [342]

Usando las relaciones entre velocidad, distancia y tiempo, se encuentra que ella condujo a una velocidad media de 90,5 km/h.

--------------------------

La velocidad es la distancia dividida por el tiempo, por lo que:

$v = \frac{d}{t}$

Total de 135,75 km, o sea, $d = 135,75$
Llego en 1,5 horas, o sea, $t = 1,5$

La velocidad es:

$v = \frac{d}{t} = \frac{135,75}{1,5}$

División de decimales, o sea, seguimos multiplicando los números por 10 hasta que ninguno sea decimal:

$v = \frac{135,75}{1,5} = \frac{1357,5}{15} = \frac{13575}{150} = 90,5$

Ella condujo a una velocidad media de 90,5 km/h.

Un problema similar es dado en brainly.com/question/24558377

4 0

1 year 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