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(1 point) Let S be ($-\infty$, 2] $\cup$ [20, $\infty$). Then S can also be described in set notation by the inequality $|x - a| \ge b$ for a = and b =

          (1 point) Let S be ($-\infty$, 2] $\cup$ [20, $\infty$).
Then S can also be described in set notation by the inequality $|x - a| \ge b$
for
a = 
and
b =
        
(1 point) Let S be (-∞, 2] ∪ [20, ∞).
Then S can also be described in set notation by the inequality |x - a| ≥ b
for
a = 
and
b =

Added by Juan Antonio P.

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Precalculus with Limits
Precalculus with Limits
Ron Larson 2nd Edition
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point) Let S be (-infty ,2]cup [20,infty ). Then S can also be described in set notation by the inequality |x-a|>=b for a= and b= Note: You can earn partial credit on this problem. Previous Problem Problem List Next Problem (1 point) Let S be (o,2] U[20,o) Then S can also be described in set notation by the inequality [ -- a| b for a= and b= Note: You can earn partial credit on this problem
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00:01 Now, solution of this question, so given that f of x y z is equal to 2x plus 5y plus 5z, this is equation 1 and subject x square plus y square plus z square is equal to 16, this is equation 1 and this is equation 2.
00:22 So, f of x this and this is g of x y of z, this is equal to x square plus y square plus z square minus 16.
00:36 So, df is equal to df plus lambda dy and df this is equal to 2dx plus 5dy plus 5dz plus lambda 2x dx plus 2y dy plus 2z dz.
01:02 So, df is equal to 0.
01:04 Now, put df is equal to 0.
01:08 Now, put df is equal to 0, then 2 plus 2x lambda is equal to 0 this is equation 3 and this is 5 plus 2y lambda is equal to 0 and 5 plus 2z lambda is equal to 0.
01:24 This is equation 4 and this is equation 5...
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