How did you find six things in triangle

What is the slope of a line that passes through (-4,-13) and (19,11)

inysia [295]

The slope of the line is the gradient, which you can find through rise over run

m (gradient) = (y1 - y2) / (x1 - x2)

where (x1, y1) is the coordinate of the first point, and (x2, x2) is the coordinate of the second point

in your question:
x1 = -4
x2 = 19
y1 = -13
y2 = 11

m = (-13 -11) / (-4 -19) = -24 / -23 = 24/23 or 1.04 (2d.p.)

hope that helps :)

3 0

3 years ago

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Find the missing part of the right triangle

masya89 [10]

The missing part is the number

3 0

3 years ago

Read 2 more answers

HELP HELP HELP HELP HELP HELL

vitfil [10]

Answer:

it sould be D because rise over run.

Step-by-step explanation:

at the point its 5/6 and the only answer with that fraction is D. or it depends because if you're looking for the begening of the dotted lines then it's C .but if you're looking for end point then it's D. if you don't know ask me and ill see what i can do

3 0

3 years ago

PLEASE ANSWER HONESTLY (PLEASE BE HELPFULL I HAVE BEEN STUCK ON THIS PROBLEM) WILL MARK BRANIEST IF CORRECT. (no links my comput

tankabanditka [31]

Answer:

a^2+b^2=c^2 look below

Step-by-step explanation:

Count the squares so

3^2+4^2=5^2

3*3=9

4*4=16

5*5=25

9+16=25 and it does check out

this should help if u need more explanation ill be more then happy to explain to you

7 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