Hello,
y=k*(x-2)(x-4)
and is passing throught (3,1)
==>1=k*(3-2)(3-4)==>k=-1
y=-(x-2)(x-4) is an answer
$24.75
this is because 22 ÷ 8 × 9 is equal to 24.75
this is the correct answer
The first and crucial thing we want to take notice is that the lines DF and EG intersect at point H which creates 4 different triangles in the rhombus. The innermost angles of DGH and EFH are vertical angles and vertical angles are congruent. So if the angles of the triangles are congruent than the triangles themselves are congruent. This is supported by the vertical angles theorem.
<span>Answer:
n=64 is large enough to use a z-test. The two-tailed 90% confidence interval (5% in each tail) is
pop. mean +/- 1.64 (s.d. / sqrt(n) ) = 16 +/- 1.64 * 0.2/8 oz = [15.959, 16.041]oz
m</span>
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:
