-2
-2
Rewrite as, int[sinx 1/2 ] = - (2/3)cosx 3/2 + C ==================or = - (2/3)sqrt[cosx 3] + C ==================
-2
a^2+b^2=c^2 3^2+4^2=c^2 9+16=c^2 25=c^2 (take square root of both sides) c=5
−3°C > 2°C −3°C < 2°C 3°C > −2°C 3°C < −2°c one of theses
Integral of sqrt(2x) = integral of (2x)1/2 = √2/(3/2)*(x)3/2 + c = (2√2)/3*(x)3/2 + c where c is the constant of integration. Check: ( (2√2)/3*(2x)3/2 + c )' = (2√2)/3*(3/2)(x)(3/2)-1 + 0 = √2*(x)1/2 = √2x
-c + 2 = 5 -c = 3 c = -3
-2
A B C + 1 2 3= 357
-2
Rewrite as, int[sinx 1/2 ] = - (2/3)cosx 3/2 + C ==================or = - (2/3)sqrt[cosx 3] + C ==================
y = mx + c Here y = -4x + c Since it passes through -2, 3 3 = (-4)(-2) + c 3 = 8 + c 3 - 8 = c c = -5 Hence equation is y = -4x -5
-2
a+b+c=4 a^2+b^2+c^2=10 a^3+b^3+c^3=22 Hope that helps. :)
a^2+b^2=c^2 3^2+4^2=c^2 9+16=c^2 25=c^2 (take square root of both sides) c=5
5 - c = 3 c = 5 - 3 Therefore, c = 2