Answer:
Equation of tangent plane to given parametric equation is:
Step-by-step explanation:
Given equation
---(1)
Normal vector tangent to plane is:
Normal vector tangent to plane is given by:
![r_{u} \times r_{v} =det\left[\begin{array}{ccc}\hat{i}&\hat{j}&\hat{k}\\cos(v)&sin(v)&0\\-usin(v)&ucos(v)&1\end{array}\right]](https://tex.z-dn.net/?f=r_%7Bu%7D%20%5Ctimes%20r_%7Bv%7D%20%3Ddet%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%5Chat%7Bi%7D%26%5Chat%7Bj%7D%26%5Chat%7Bk%7D%5C%5Ccos%28v%29%26sin%28v%29%260%5C%5C-usin%28v%29%26ucos%28v%29%261%5Cend%7Barray%7D%5Cright%5D)
Expanding with first row
at u=5, v =π/3
---(2)
at u=5, v =π/3 (1) becomes,
From above eq coordinates of r₀ can be found as:
From (2) coordinates of normal vector can be found as
Equation of tangent line can be found as:
P(2 aces) = (1/13)^2
P(2 kings) = (1/13)^2
P(king and ace) = 8C2/52C2 - 2(1/13)^2 = 0.0152
Answer:
log5(125)=3 The 5 should be smaller and a little lower.
Step-by-step explanation:
Answer:
d. 8°
Step-by-step explanation:
It appears as though you intend ∠ECF and ∠ACB to be vertical angles, hence the same measure, 47°. The angle of interest, ∠BCD, is added to that to make ∠ACD, which is 55°. The added angle must be 55° -47° = 8°.
Answer:
2.33
Step-by-step explanation:
imagine a circle. its center is A, and it goes through B, so its radius is AB.
then it is important to know that the sum of all the angles in a triangle is 180 degrees.
one angle (at C) is 90. the angle at B is 25. so, the angle at A is 180 - 90 - 25 = 65 degrees.
more back to our circle.
in this circle the line CB is the sine of the angle at A multiplied by the radius.
and AC is the cosine of the angle at A multiplied by the radius.
we can ignore the orientation + and - of these functions, as we are only interested in the absolute length (and we can mirror the triangle, and all the angles and side lengths still stay the same).
=> CB = sin(A)×AB
AC = cos(A)×AB
=> 5 = sin(65)×AB
=> AB = 5 / sin(65)
=> AC = cos(65)×5/sin(65) = 5 × (cos(65)/sin(65)) =
= 5 × cot(65) = 2.33
