Hey there,
The answer is true
Hope it helped!
Answer:
16
Step-by-step explanation:
Since line CE is equal to 32 and line DB is the bisector (cut into the middle) of line CE then line CF would be half of line CE. Hence the reason that line CF is equal to 16
<span>Maximum area = sqrt(3)/8
Let's first express the width of the triangle as a function of it's height.
If you draw an equilateral triangle, then a rectangle using one of the triangles edges as the base, you'll see that there's 4 regions created. They are the rectangle, a smaller equilateral triangle above the rectangle, and 2 right triangles with one leg being the height of the rectangle and the other 2 angles being 30 and 60 degrees. Let's call the short leg of that triangle b. And that makes the width of the rectangle equal to 1 minus twice b. So we have
w = 1 - 2b
b = h/sqrt(3)
So
w = 1 - 2*h/sqrt(3)
The area of the rectangle is
A = hw
A = h(1 - 2*h/sqrt(3))
A = h*1 - h*2*h/sqrt(3)
A = h - 2h^2/sqrt(3)
We now have a quadratic equation where A = -2/sqrt(3), b = 1, and c=0.
We can solve the problem by using a bit of calculus and calculating the first derivative, then solving for 0. But since this is a simple quadratic, we could also take advantage that a parabola is symmetrical and that the maximum value will be the midpoint between it's roots. So let's use the quadratic formula and solve it that way. The 2 roots are 0, and 1.5/sqrt(3).
The midpoint is
(0 + 1.5/sqrt(3))/2 = 1.5/sqrt(3) / 2 = 0.75/sqrt(3)
So the desired height is 0.75/sqrt(3).
Now let's calculate the width:
w = 1 - 2*h/sqrt(3)
w = 1 - 2* 0.75/sqrt(3) /sqrt(3)
w = 1 - 2* 0.75/3
w = 1 - 1.5/3
w = 1 - 0.5
w = 0.5
The area is
A = hw
A = 0.75/sqrt(3) * 0.5
A = 0.375/sqrt(3)
Now as I said earlier, we could use the first derivative. Let's do that as well and see what happens.
A = h - 2h^2/sqrt(3)
A' = 1h^0 - 4h/sqrt(3)
A' = 1 - 4h/sqrt(3)
Now solve for 0.
A' = 1 - 4h/sqrt(3)
0 = 1 - 4h/sqrt(3)
4h/sqrt(3) = 1
4h = sqrt(3)
h = sqrt(3)/4
w = 1 - 2*(sqrt(3)/4)/sqrt(3)
w = 1 - 2/4
w = 1 -1/2
w = 1/2
A = wh
A = 1/2 * sqrt(3)/4
A = sqrt(3)/8
And the other method got us 0.375/sqrt(3). Are they the same? Let's see.
0.375/sqrt(3)
Multiply top and bottom by sqrt(3)
0.375*sqrt(3)/3
Multiply top and bottom by 8
3*sqrt(3)/24
Divide top and bottom by 3
sqrt(3)/8
Yep, they're the same.
And since sqrt(3)/8 looks so much nicer than 0.375/sqrt(3), let's use that as the answer.</span>
Since the bucket has already a mass of 2,000 kg, therefore
the maximum mass of gravel would only be:
15,000 kg – 2,000 kg = 13,000 kg gravel
Solving for volume:
volume = 13,000 kg / (1,500 kg / 1 m^3)
volume = 8.67 m^3
<span>Therefore about 8.67 cubic meter of gravel can be safely
loaded by the crane.</span>
Answer:
Step-by-step explanation:
For two matrix P and Q, the product, say PQ is defined when:
The number of columns of P = The number of rows of Q
Since A is a 2×2 matrix and B is also a 2×2 matrix
Thus both AB and BA are possible.
So AB is:
BA is:
