Step-by-step explanation:
i)=
m=2/3
x=-3
y = 5
now
y=mx+c
or, 5 = 2/3× -3 + c
or, 5 = -2+c
c =7
now,
y= mx + c
or, y = 2/3×x+7
or, y = 2x/3+7
or, 3y = 2x+21
so, 3y-2x=21
equation would be 3y-2x=21.
now
ii)=
if slope of 3y-2x=21 and 2x-3y = 0 are same, they are parallel.
2x-3y = 0
2x = 3y
lets suppose x is 9
now
2x=3y
2×9 = 3y
or, 18 = 3y
so, y = 6
one coordinate would be (9,6) now
lets again suppose x is 3
2x=3y
2×3 = 3y
or, 6 = 3y
so, y = 2
another coordinate is (3,2)
now lets find slope
x1=9,y1=6
x2=3,y2=2
slope = (y2-y1)/(x2-x1)
= (2-6)/(3-9)
= -4/-6
= 2/3
Since, slope of 3y-2x=21 and 2x-3y has same slope 2/3, they are parallel.
<span>5(3x^3)^3-2(3x^3)
=</span><span>5(3^3 x^9) - 6x^3
= 135x^9 - 6x^3
= 3x^3 (45x^6 - 2)</span>
A) The complement of an angle is the other angle that can be added to the original to add up to 90 degrees. The complement of <ABD would be 90-36
=54
b) The supplement of an angle is whatever number can be added to the original to add up to 180 degrees. The supplement of <ABD would be 180-36
=144
Answer:
B.
C.
D.
E.
Step-by-step explanation:
Area of the parabolic region = Integral of [a^2 - x^2 ]dx | from - a to a =
(a^2)x - (x^3)/3 | from - a to a = (a^2)(a) - (a^3)/3 - (a^2)(-a) + (-a^3)/3 =
= 2a^3 - 2(a^3)/3 = [4/3](a^3)
Area of the triangle = [1/2]base*height = [1/2](2a)(a)^2 = <span>a^3
ratio area of the triangle / area of the parabolic region = a^3 / {[4/3](a^3)} =
Limit of </span><span><span>a^3 / {[4/3](a^3)} </span>as a -> 0 = 1 /(4/3) = 4/3
</span>
