The area of the surface is 144.708
The equation of the given surface is,
z=g(x,y)=xy
Solving the partial derivatives,
∂g∂x=y,∂g∂y=x
Substituting to the formula
S=∬√1+( ∂g∂x)2+(∂g∂y)2dA
Thus,
S=∬√1+(y)2+(x)2dAS=∬√1+x2+y2dA
The region in the XY-plane is defined by the intervals 0≤θ≤2π,0≤r≤4
Converting the integral into polar coordinates,
S=∫2π0∫40√1+r2rdrdθ
Integrating with respect to r
S=∫2π0[13(1+r2)32]40dθ
S=∫2π0(17√173−13)dθ
Integrating with respect to θ
S=(17√173−13)[θ]2π0
S≈144.708
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Answer:
12:3
Step-by-step explanation:
<span> x − 2y = −1 => x = 2y -1
2x + y = −12
2(2y - 1) + y = -12
4y - 2 + y = -12
5y = -10
y = -2
x = 2</span>·(-2) - 1
x = -4 - 1
x = -5
Answer <span>D (-5,-2)</span>
This fraction cannot be turned into a mixed number but it can be reduced to

or 0.75
hope this helps
Using a discrete probability distribution, it is found that:
a) There is a 0.3 = 30% probability that he will mow exactly 2 lawns on a randomly selected day.
b) There is a 0.8 = 80% probability that he will mow at least 1 lawn on a randomly selected day.
c) The expected value is of 1.3 lawns mowed on a randomly selected day.
<h3>What is the discrete probability distribution?</h3>
Researching the problem on the internet, it is found that the distribution for the number of lawns mowed on a randomly selected dayis given by:
Item a:
P(X = 2) = 0.3, hence, there is a 0.3 = 30% probability that he will mow exactly 2 lawns on a randomly selected day.
Item b:

There is a 0.8 = 80% probability that he will mow at least 1 lawn on a randomly selected day.
Item c:
The expected value of a discrete distribution is given by the <u>sum of each value multiplied by it's respective probability</u>, hence:
E(X) = 0(0.2) + 1(0.4) + 2(0.3) + 3(0.1) = 1.3.
The expected value is of 1.3 lawns mowed on a randomly selected day.
More can be learned about discrete probability distributions at brainly.com/question/24855677