What side is this a isosceles a scalene or equilateral

Complete the point slope equation of the line through (-5, 7) and (-4, 0).

IgorLugansk [536]

For this case we have that by definition, the point-slope equation of a line is given by:

$y = mx + b$

Where:

m: It's the slope

b: It is the cut-off point with the y axis

We have two points:

$(x_ {1}, y_ {1}): (- 5,7)\\(x_ {2}, y_ {2}): (- 4,0)$

We found the slope:

$m = \frac {y_ {2} -y_ {1}} {x_ {2} -x_ {1}} = \frac {0-7} {- 4 - (- 5)} = \frac {-7} {-4 + 5} = \frac {-7} {1} = - 7$

Thus, the equation is of the form:

$y = -7x + b$

We substitute one of the points and find "b":

$0 = -7 (-4) + b\\0 = 28 + b\\b = -28$

Finally, the equation is:

$y = -7x-28$

Answer:

$y = -7x-28$

5 0

3 years ago

Evaluate the surface integral ModifyingBelow Integral from nothing to nothing Integral from nothing to nothing With Upper S f (x

kykrilka [37]

Parameterize $S$ by

$\vec s(u,v)=6\cos u\sin v\,\vec\imath+6\sin u\sin v\,\vec\jmath+6\cos v\,\vec k$

with $0\le u\le2\pi$ and $0\le v\le\frac\pi2$ . Take a normal vector to $S$ ,

$\dfrac{\partial\vec s}{\partial v}\times\dfrac{\partial\vec s}{\partial u}=36\cos u\sin^2v\,\vec\imath+36\sin u\sin^2v\,\vec\jmath+36\cos v\sin v\,\vec k$

which has norm

$\left\|\dfrac{\partial\vec s}{\partial v}\times\dfrac{\partial\vec s}{\partial u}\right\|=36\sin v$

Then the integral of $f(x,y,z)=x^2+y^2$ over $S$ is

$\displaystyle\iint_Sx^2+y^2\,\mathrm dS=\iint_S\left((6\cos u\sin v)^2+(6\sin u\sin v)^2\right)\left\|\frac{\partial\vec s}{\partial v}\times\frac{\partial\vec s}{\partial u}\right\|\,\mathrm du\,\mathrm dv$