What is the area of the triangular base?
8 square feet
What is the length of edge DF?
4 feet
Answer:
dy/dx = -1/√(1 - x²)
For 0 < y < π
Step-by-step explanation:
Given the function cos y = x
-siny dy = dx
-siny dy/dx = 1
dy/dx = -1/siny (equation 1)
But cos²y + sin²y = 1
=> sin²y = 1 - cos²y
=> siny = √(1 - cos²y) (equation 2)
Again, we know that
cosy = x
=> cos²y = x² (equation 3)
Using (equation 3) in (equation 2), we have
siny = √(1 - x²) (equation 4)
Finally, using (equation 4) in (equation 1), we have
dy/dx = -1/√(1 - x²)
The largest interval is when
√(1 - x²) = 0
=> 1 - x² = 0
=> x² = 1
=> x = ±1
So, the interval is
-1 < x < 1
arccos(1) < y < arxcos(-1)
= 0 < y < π
Answer:
Quadrant I and III
Step-by-step explanation:
The coordinate (3,9) is all positive, therefore it lies in quadrant I.
The coordinate (-3,-9) is all negative, therefore it lies in quadrant III.
To solve this problem you must keep on mind the following information: By definition, a quadratic function has the following form:
Where
is the leading coefficient.
If the leading coeficient is closer to zero, the parabola is widest,if it has a larger positive or negative value, the parabola is narrowest.
Therefore, by knowing the information above, you have that
the answer is: 
Answer:
The parametric equations for the tangent line are
:
x = Cos(10) - t×Sin(10)
y = Sin(10) + t×Cos(10)
z = 20 + 2t
Step-by-step explanation:
When Z=20:
Z=2t=20 ⇒ t=10
The point of tangency is:
r(10)= Cos(10) i + Sin(10) j + 20 k
We have to find the derivative of r(t) to get the tangent line:
r'(t)= -Sin(t) i + Cos(t) j + 2 k
The direction vector at t=10 is:
r'(10)= -Sin(10) i + Cos(10) j + 2 k
So, the equation of the tangent line is given by:
x = cos 10 -t×Sin(10)
y = sin 10 + t×Cos(10)
z = 20 + 2t
