Answer:
Step-by-step explanation:
Part A
Since, the given angles are formed at a point are on a straight line, sum of these angles will be 180°.
Part B
(3x - 5)° + (4x + 2)° + (2x + 3)° = 180°
9x = 180
x = 20
Part C
m∠ADB = (3x - 5)°
= (3×20 - 5)°
= 55°
m∠DBK = (4x + 2)°
= (4×20 + 2)°
= 82°
m∠KBC = (2x + 3)°
= (2×20 + 3)°
= 43°
(m + 3)^2 -16
=(m + 3)^2 - 4^2
=((m+3) + 4)*((m+3) - 4)
=(m+7)(m-1)
Here are the formulas i used:
so first i made 16 to 4^2, then i use (a^2 -b^2)= (a+b)(a-b) . In this case a would be (m+3) and b is 4.
Let ????C be the positively oriented square with vertices (0,0)(0,0), (2,0)(2,0), (2,2)(2,2), (0,2)(0,2). Use Green's Theorem to
bonufazy [111]
Answer:
-48
Step-by-step explanation:
Lets call L(x,y) = 10y²x, M(x,y) = 4x²y. Green's Theorem stays that the line integral over C can be calculed by computing the double integral over the inner square of Mx - Ly. In other words
Where Mx and Ly are the partial derivates of M and L with respect to the x variable and the y variable respectively. In other words, Mx is obtained from M by derivating over the variable x treating y as constant, and Ly is obtaining derivating L over y by treateing x as constant. Hence,
- M(x,y) = 4x²y
- Mx(x,y) = 8xy
- L(x,y) = 10y²x
- Ly(x,y) = 20xy
- Mx - Ly = -12xy
Therefore, the line integral can be computed as follows
Using the linearity of the integral and Barrow's Theorem we have
As a result, the value of the double integral is -48-