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3 years ago

Reflect the x axis A B C D

Mathematics

1 answer:

Ronch [10]3 years ago

3 0

Answer:

see explanation

Step-by-step explanation:

Under a reflection in the x- axis

a point (x, y ) → (x, - y ) , then

A (- 1, - 17 ) → A' (- 1, 17 )

B (0, - 12 ) → B' (0, 12 )

C (- 5, - 11 ) → C' (- 5, 11 )

D (- 6, - 16 ) → D' (- 6, 16 )

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Please help how to solve

tatiyna

Answer:

$= > in \: the \: bigger \: triangle \\ height = 14 \: ft \\ base = 22.5 \: ft \\ = > in \: smaller \: triangle \\ height = x \: ft \\ base = 9 \: ft \\ using \: ratios \\ \frac{bigger \: height}{smaller \: height} = \frac{bigger \: base}{smaller \: base} \\ \frac{14}{x} = \frac{22.5}{9} \\ \frac{14}{x} = 2.5 \\ x = \frac{14}{2.5} \\ x = 5.6 \: ft$

7 0

3 years ago

CAN SOMEONE PLEASE HELP ME WITH THIS PROBLEM

Klio2033 [76]

Answer:

V ≈ 2011 cm³

Step-by-step explanation:

The volume (V) of a cylinder is calculated as

V = area of base × perpendicular height

V = πr²h ( r is the radius and h the height )

Here r = 8 and h = 10, thus

V = π × 8² × 10

= π × 64 × 10 = 640π ≈ 2011 cm³ ( to nearest whole number )

3 0

3 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)