All categories

3 years ago

2 (01.05)

Mathematics

1 answer:

lord [1]3 years ago

3 0

Answer:

y +2 = 4(x -3)

Step-by-step explanation:

Point-slope form of the equation of a line with slope m through point (h, k) is ...

y -k = m(x -h)

For m = 4 and (h, k) = (3, -2), the equation is ...

y +2 = 4(x -3) . . . . . matches the first choice

You might be interested in

Which expression is equivalent to (x Superscript 27 Baseline y) Superscript one-third?

alexgriva [62]

Answer:

x Superscript 9 Baseline (RootIndex 3 StartRoot y EndRoot)

OR x^9/(∛y)

Step-by-step explanation:

Given the indicinal equation

(x^27/y)^1/3

To find the corresponding expression, we will simplify the equation as shown:

(x^27/y)^⅓

= (x^27)^⅓/y⅓

= {x^(3×9)}^⅓/y⅓

= x^9/y⅓

= x^9/(∛y)

The right answer is x Superscript 9 Baseline (RootIndex 3 StartRoot y EndRoot)

6 0

3 years ago

Read 2 more answers

A baseball hat for $22.90 at 5% sales tax

Alexus [3.1K]

22.90*1.05=24.045
sales tax is 5% added to 100% of the original cost making the actaul cost including tax $24.045

8 0

3 years ago

Read 2 more answers

A bag of marbles contains 3 yellow, 2 red, 2 green, and 1 blue marble. What is the probability of selecting a green marble?

Papessa [141]

Answer:

2/8

Step-by-step explanation:

There are 2 green marbles.

In all, there are 8 marbles.

So, the probability is 2/8.

6 0

3 years ago

Read 2 more answers

Ricky says (x5)(x3) = x15 Rocky says (x5)(x) = x ? Who is correct? Why?

atroni [7]

Answer:

1.rocky

Step-by-step explanation

1.we have the add the powers of x,not multiply.

6 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)

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