we know that
The conjugate of a complex number is the number with equal real part and imaginary part equal in magnitude but opposite in sign
so
In this problem we have
the conjugate is equal to--------> 
therefore
<u>the answer is the option</u>
10+3i
I am assuming the 32 cases contain the small cans...
32 cases...with 28 small cans of fruit in each case..
32 * 28 = 896 small cans of fruit in total
24 cases...with 13 large cans of fruit in each case...
24 * 13 = 312 large cans of fruit in total
so there are (896 - 312) = 584 more small cans then large cans <==
Answer:
<h2>A = 225 in²</h2>
Step-by-step explanation:
The formula of an area of a square:
<em>s</em><em> - side</em>
The formula of a perimeter of a square:
<em>s</em><em> - side</em>
<em />
We have a perimeter
Substitute:
<em>divide both sides by 4</em>
Put it to the formula of an area of a square:
60, 120, 180
It’s honestly super easy each hour is 69 multiply it by 2 then bultiply 60 by 3
The dimensions of the enclosure that is most economical to construct are; x = 14.22 ft and y = 22.5 ft
<h3>How to maximize area?</h3>
Let the length of the rectangular area be x feet
Let the width of the area = y feet
Area of the rectangle = xy square feet
Or xy = 320 square feet
y = 320/x -----(1)
Cost to fence the three sides = $6 per foot
Therefore cost to fence one length and two width of the rectangular area
= 6(x + 2y)
Similarly cost to fence the fourth side = $13 per foot
So, the cost of the remaining length = 13x
Total cost to fence = 6(x + 2y) + 13x
Cost (C) = 6(x + 2y) + 13x
C = 6x + 12y + 13x
C = 19x + 12y
From equation (1),
C = 19x + 12(320/x)
C' = 19 - 3840/x²
At C' = 0, we have;
19 - 3840/x² = 0
19 = 3840/x²
19x² = 3840
x² = 3840/19
x = √(3840/19)
x = 14.22 ft
y = 320/14.22
y = 22.5 ft
Read more about Maximization of Area at; brainly.com/question/13869651
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