Answer:
2/8 Green
Step-by-step explanation:
Since the green has two sections on the spinner it has two chances out of 8 to be landed on.
Answer:
D. 1.3
Step-by-step explanation:
Perimeter (square) = 4*side
4 = 4*side
side = 1
We have triangle QVR. QV is a hypotenuse.
If we look at the triangle QTR, QT is a hypotenuse.
QV should be less than QT.
Let's find QT using Pythagorean theorem.
c² = a² + b²
QT² = QR² + RT² = 1² + 1² = 2
QT = √2 ≈ 1.4
QV < QT
QV < 1.4
We have 1.3 and 1 that are less than 1.4, but 1 is a length of a leg of the triangle, so 1 is impossible.
Answer D. 1.3
<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>
Answer:
y = -3x + 7
Step-by-step explanation:
when they are parallel, they always have the same slope.
