Simplify the expression. 6p5

1 answer:

3 0

If you are considering P of the permutation and combination. Then simply pit the figure 6P5 in calculator and your answer will be 720. Or if not calculator then the formula is P(n,r)= n(factorial)/ (n-r)factorial.

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Find the absolute maximum and minimum values of the following function in the closed region bounded by the triangle with vertice

alukav5142 [94]

$f(x,y)=2x^2-8x+y^2-8y+2$

$f_x(x,y)=4x-8=0\implies x=2$

$f_y(x,y)=2y-8=0\implies y=4$

There is one critical point at (2, 4), but this point happens to fall on one of the boundaries of the region. We'll get to that point in a moment.

Along the boundary $x=0$ , we have

$f(0,y)=y^2-8y+2=(y-4)^2-14$

which attains a maximum value of

$f(0,4)=-14$

Along $y=44$ , we have

$f(x,44)=2x^2-8x+1586=2(x-2)^2+1578$

which attains a maximum of

$f(2,44)=1578$

Along $y=2x$ , we have

$f(x,2x)=6x^2-24x+2=6(x-2)^2-22$

which attains a maximum of

$f(2,4)=-22$

So over the given region, the absolute maximum of $f(x,y)$ is 1578 at (2, 44).

8 0

2 years ago

Quadrilateral KLMN is a rectangle. The coordinates of L are L(1,-4) and the coordinates of M are M(3,-2) find the slopes of side

m_a_m_a [10]

We can start by finding the gradient of LM
$m= \frac{ y_{2}- y_{1} }{x_{2} - x_{1} } = \frac{-2--4}{3-1} = \frac{2}{2}=1$

Two perpendicular lines will meet the requirement $m_{1}$ × $m_{2}$ =-1
Two parallel lines have equal gradients

NM is perpendicular to LM, hence the gradient of NM is -1
KN is a line that is parallel to NM, hence the gradient is 1
KL is perpendicular to LM, hence the gradient of KL is -1