Answer:
option B
x = 1 + 3√6 or x = 1 - 3√6
Step-by-step explanation:
Given in the question an equation,
3(x-1)² - 162 = 0
rearrange the x terms to the left and constant to the right
3(x-1)² = 162
(x-1)² = 162/3
(x-1)² = 54
Take square root on both sides
√(x-1)² = √54
x - 1 = ±3√6
x = ±3√6 + 1
So we have two values for x
x = 3√6 + 1 OR x = -3√6 + 1
60 cupcakes because u can put 5 rows with 12 in each or u can do 12 rows with 5 in each.
Find all the side lengths and angle measures using trig (law of sines and cosines, assuming they are all only one-triangle possibilities).
The perimeter is 6x + 12, and a model example is attached.
All three sides of an equilateral triangle are the same, so multiply the side length by 3 to find the perimeter.
3(2x + 4)
Now distribute the 3.
3 * 2x + 3 * 4
6x + 12
The maximum volume of the box is 40√(10/27) cu in.
Here we see that volume is to be maximized
The surface area of the box is 40 sq in
Since the top lid is open, the surface area will be
lb + 2lh + 2bh = 40
Now, the length is equal to the breadth.
Let them be x in
Hence,
x² + 2xh + 2xh = 40
or, 4xh = 40 - x²
or, h = 10/x - x/4
Let f(x) = volume of the box
= lbh
Hence,
f(x) = x²(10/x - x/4)
= 10x - x³/4
differentiating with respect to x and equating it to 0 gives us
f'(x) = 10 - 3x²/4 = 0
or, 3x²/4 = 10
or, x² = 40/3
Hence x will be equal to 2√(10/3)
Now to check whether this value of x will give us the max volume, we will find
f"(2√(10/3))
f"(x) = -3x/2
hence,
f"(2√(10/3)) = -3√(10/3)
Since the above value is negative, volume is maximum for x = 2√(10/3)
Hence volume
= 10 X 2√(10/3) - [2√(10/3)]³/4
= 2√(10/3) [10 - 10/3]
= 2√(10/3) X 20/3
= 40√(10/27) cu in
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