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KIM [24]
3 years ago
13

-3 (5-b)= Use the distributive property to rewrite this expression

Mathematics
2 answers:
ollegr [7]3 years ago
7 0
= -3 (5-b) 
= -3 (5 + (-b))
= (-3 x -b) + (-3 x 5)
= 3b + (-15)
= 3b -15
Brums [2.3K]3 years ago
5 0
-3 ( 5 - b ) =
-15 + 3b = 
That is how you rewrite it, because it means the same thing.
-3 ( 5 - b ) = 
-15 + 3b = 
+15            +15
3b = 15
/3      /3
b = 5 
incase you wanted to know how to solve to.
 I hope this helped, HAVE A NICE DAY.
WELCOME :/
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Looks like f(x,y)=x^2+y^2-x^2y+7.

f_x=2x-2xy=0\implies2x(1-y)=0\implies x=0\text{ or }y=1

f_y=2y-x^2=0\implies2y=x^2

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The latter two critical points occur outside of D since |\pm\sqrt2|>1 so we ignore those points.

The Hessian matrix for this function is

H(x,y)=\begin{bmatrix}f_{xx}&f_{xy}\\f_{yx}&f_{yy}\end{bmatrix}=\begin{bmatrix}2-2y&-2x\\-2x&2\end{bmatrix}

The value of its determinant at (0, 0) is \det H(0,0)=4>0, which means a minimum occurs at the point, and we have f(0,0)=7.

Now consider each boundary:

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f(1,y)=8-y+y^2=\left(y-\dfrac12\right)^2+\dfrac{31}4

which has 3 extreme values over the interval -1\le y\le1 of 31/4 = 7.75 at the point (1, 1/2); 8 at (1, 1); and 10 at (1, -1).

  • If x=-1, then

f(-1,y)=8-y+y^2

and we get the same extrema as in the previous case: 8 at (-1, 1), and 10 at (-1, -1).

  • If y=1, then

f(x,1)=8

which doesn't tell us about anything we don't already know (namely that 8 is an extreme value).

  • If y=-1, then

f(x,-1)=2x^2+8

which has 3 extreme values, but the previous cases already include them.

Hence f(x,y) has absolute maxima of 10 at the points (1, -1) and (-1, -1) and an absolute minimum of 0 at (0, 0).

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