We get domain from X-axis:
<h2>
0 ≤ x < ∞</h2>
Answer:
0
Step-by-step explanation:
If we use polar coordinates, the region D can be covered by replacing (x,y) by (r*sin(Θ),rcosΘ)), with 0<r<7, 0<Θ<2π. The differential matrix
![\left[\begin{array}{cc}rcos(\theta)&-rsin(\theta)\\sin(\theta)&cos(\theta)\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Drcos%28%5Ctheta%29%26-rsin%28%5Ctheta%29%5C%5Csin%28%5Ctheta%29%26cos%28%5Ctheta%29%5Cend%7Barray%7D%5Cright%5D)
has determinant equal to r, so we can compute the double integral as follows
(Note that we multiplied by the determinant of the Jacobian, r). A primitive for r³ is r⁴/4, thus, for Barrow's rule we have
A primitive of cos(Θ)sin(Θ) can be obtained using substitution, and it is sin²(Θ)/2 (note that the derivate of sin²(Θ) is 2sin(Θ)cos(Θ)). Therefore, taking both the dividing 4 and the 2 obtained, we have
Hence, the integral is 0.
Answer:
Step-by-step explanation: (a) y = 3x - 8 (b) 3y + x - 16 = 0
(a) The line is y = 3x - 2
But the condition for parallelism is that for two lines to be parallel to each other, their gradients m must be equal, ie, m1 = m2
therefore, the gradient of the line above m1 = 3, m2 = 3
since the line passes through the coordinate of ( 4, 4 ),
we need to find the y intersect ( c ) by substitute for x, m and y in the equation below.
y = mx + c
4 = 3 x 4 + c
4 = 12 + c
c = 4 - 12
c = -8
Therefore, substitute for c in the equation of a line above to get the second equation
y = mx + c
y = 3x - 8
(b) Condition for perpendicularity of two line is that the product of their gradients must be( -1 )
ie, m1m2 = -1
Now from the equation above, y = 3x - 2, m1 = 3 and m2 = -1/3
to get the value of c, we substitute for x, y and m into the equation
y = mx + c
4 = -1/3 x 4 = c
4 = -4/3 + c
multiply through by 3 to make it a linear equation
12 = -4 + 3c
12 + 4 = 3c
16 = 3c
c = 16/3
now put c = 16/3 into the equation , y=mx + c
y = -x/3 + 16/3
multiply through by 3
3y = -x + 16
3y + x - 16 =0
Male : female : total
5 : 7 : 5 + 7
5: 7 : 12
12600 ÷ 12 = 1050
1050 x 5 = 5250 male students
1050 x 7 = 7350 female students
