Multiply by 0.4 on both sides to cancel the original 0.4 and change the 10
<span>Add both the result (1×3 + 1×3) = 6 and write down to the left of 9 (result ofstep 1).
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Using BEDMAS
a) 6+4(2+3)^2
6+4(5)^2 - Eliminate addition in bracket
6+4(25) - Execute the exponent
6+100 - Mulitplying
106 - Addition
b)(6+4)(2+3)
(10)(5) -Addition in BRACKETS done
50 - Mulipication
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Answer:
ans: it's order wise
a) yes
b) no
c) no
d) no
Step-by-step explanation:
use pythagoras theorem
h² = p² + b²
data given satisfying the theorem form right angled triangle
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
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