How many positive two-digit numbers are evenly divisible by 6

What is the modulus after (-1+sqrt3i) gets raised to the 6th power

Eddi Din [679]

Answer:

it is b

Step-by-step explanation:

took the test

8 0

3 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)

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

When full, a cinema house has n people. When 280 people have taken seat the remaining seats comes to $384 if each tickets cost $

Yuki888 [10]

I don’t understand the question, can you restate it?

5 0

3 years ago

Read 2 more answers

The change in the value of an account when given to the nearest dollar ? Identify the set of numbers that best describes .

AveGali [126]

When an account is rounded off to its nearest dollar will change based on the place value. For example, the given account of money if $ 56 730, then it will be rounded to its nearest ten thousands. Then the given amount will be rounded to $57,000 in where the given money rounded up and gain up to $300. But if the money will be rounded with its nearest hundreds, the money will become $ 56,700. Notice that the given money rounded down and lost $30 dollars.

7 0

4 years ago

At the city museum, child admission is $6.20 and adult admission is $9.50. On Friday, tickets were sold for a total sales of $10

Nostrana [21]

Answer:

98child tickets were sold that day

Step by step Explanation:

Let the number of children= A of the number of adult

Then

6.20C + 9.50A = 1082.0

From the question it was said that four times as many adult tickets as child tickets were sold which implies that number of adult was FOUR times the children, then we can say

A= 4C

Then substitute into the above equation

6.20C + 9.50A = 1082.0

6.20C + 9.50(4C) = 1082.0

6.20C + 38C= 1082.0

44.2C =1082.0

C= 24.48

But we know that

A= 4C

A=4* 24.48

A= 97.9

Therefore ,98child tickets were sold that day

6 0

4 years ago