If x represents the width of the poster (including borders), the area of the finished poster can be written as
.. a = x*(390/(x -10) +8)
.. = 8x +390 +3900/(x -10)
Then the derivative with respect to x is
.. da/dx = 8 -3900/(x -10)^2
This is zero at the minimum area, where
.. x = √(3900/8) +10 ≈ 32.08 . . . . cm
The height is then
.. 390/(x -10) +8 = 8 +2√78 ≈ 25.66 . . . . cm
The poster with the smallest area is 32.08 cm wide by 25.66 cm tall.
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In these "border" problems, the smallest area will have the same overall dimension ratio that the borders have. Here, the poster is 10/8 = 1.25 times as wide as it is high.
I think so.. If not i am super sorry!!
180-58=122=x
Its honestly a piece of cake a straight line will <u>Always</u> have a degree measurement on one-hundred and eighty degrees
Answer:
Rs 57750
Step-by-step explanation:
width =5 m
length of first parallel road 70 m
area of first road 70×5=350 sq.m
length of second road 45 m
area of second road =45×5=225 sq.m
area of the common part of cross road with 5 m width that lies at the
center of the park 5×5 = 225 sq.m
area of roads = area of first road +area of second cross road−common area
=(350+225)−25
=575−25
=550sqm
cost of the construction per sq.m =Rs.105
total cost =550×105=Rs 57750
First, we can factor an x out of this.
We're left with x(x^2 - 16)
We can use the difference of squares theorem here to factor this further.
<h3><u>We're left with (x)(x + 4)(x - 4)</u></h3>
