How do you insert parentheses so the statement #4+10+8-9•8= -50# is true?

2 Answers

Answer:

Parenthesis aren't needed if PEMDAS is followed. However, parenthesis can be added to reinforce the need to multiply first.

Explanation:

Looking at the original, the key part is to get the 8 to multiply against the correct number so that it all works out to be #-50#. And it turns out that we need it to multiply against #-9#, so that:

#22-72=-50#

The way the equation is written right now, we'd have 4 adding to 10 adding to 8 and then subtracting out 72... which is exactly what we want. So my first thought is that parenthesis aren't needed. Without parenthesis, we get:

#4+10+8-9*8=-50#

#4+10+8-72=-50#

#-50=-50#

If we want to throw some parenthesis in here, then we'd add them so that it reads either as #(9*8)# - we need to keep the minus sign outside of the parenthesis because if it sits within them, it'll look like the #+8# needs to multiply against the result of #-9*8#. So that would look like:

#4+10+8-(9*8)=-50#

Answer:

No need for parentheses

Explanation:

This problem can be formulated as a minimum determination.

Find

Minimum of

#f(x_1,x_2,x_3) =(4 x_1 + 10 x_2 + 8 x_3 - 72 + 50)^2#

subjected to

#x_1^2= 1,x_2^2= 1,x_3^2=1#

The lagrangian is

#L(x_1,x_2,x_3,lambda_1,lambda_2,lambda_3) = f(x_1,x_2,x_3)+sum_{k=1}^3lambda_k(x_k^2-1)#

The stationary points are the solutions to

#grad L(x_1,x_2,x_3,lambda_1,lambda_2,lambda_3) = vec 0#

or

#{ (2 lambda_1 x_1 + 8 (-22 + 4 x1 + 10 x_2 + 8 x_3)=0), (2 lambda_2 x_2 + 20 (-22 + 4 x_1 + 10 x_2 + 8 x_3)=0), (2 lambda_3 x_3 + 16 (-22 + 4 x_1 + 10 x_2 + 8 x_3)=0), ( x_1^2-1), ( x_2^2-1), (x_3^2-1) :}#

The solutions

#( (x_1 = -1, x_2 = -1, x_3 = -1->1936), (x_1 = 1, x_2 = -1, x_3 = -1->1296), (x_1 = -1, x_2 = 1, x_3 = -1->576), (x_1 = 1, x_2 = 1, x_3 = -1->256), (x_1 = -1, x_2 = -1, x_3 = 1->784), (x_1 = 1, x_2 = -1, x_3 = 1->400), (x_1 = -1, x_2 = 1, x_3 = 1->64), (x_1 = 1, x_2 = 1, x_3 = 1->0))#

The last column shows the #f(x_1,x_2,x_3)# corresponding values

The solution is

#(x_1 = 1, x_2 = 1, x_3 = 1)# so no need for parentheses