What values of b satisfy 3(2b+3)^2 = 36
we have
3(2b+3)^2 = 36
Divide both sides by 3
(2b+3)^2 = 12
take the square root of both sides
( 2b+3)} =(+ /-) \sqrt{12} \\ 2b=(+ /-) \sqrt{12}-3
b1=\frac{\sqrt{12}}{2} -\frac{3}{2}
b1=\sqrt{3} -\frac{3}{2}
b2=\frac{-\sqrt{12}}{2} -\frac{3}{2}
b2=-\sqrt{3} -\frac{3}{2}
therefore
the answer is
the values of b are
b1=\sqrt{3} -\frac{3}{2}
b2=-\sqrt{3} -\frac{3}{2}
Answer:
Kindly check explanation
Step-by-step explanation:
Given that :
Total amount to spend = $55
Amount of food and drinks purchased = $14.25
Amount put on gaming card = $(55 - 14.25) = $40.75
Cost per game = $1.25
Number of games (g) he can play
Number of games g:
Amount put on gaming card / cost per game
= $40.75 / $1.25
= 32.6 games
g ≤ 32
Answer:
The wedge cut from the first octant ⟹ z ≥ 0 and y ≥ 0 ⟹ 12−3y^2 ≥ 0 ⟹ 0 ≤ y ≤ 2
0 ≤ y ≤ 2 and x = 2-y ⟹ 0 ≤ x ≤ 2
V = ∫∫∫ dzdydx
dz has changed from zero to 12−3y^2
dy has changed from zero to 2-x
dx has changed from zero to 2
V = ∫∫∫ dzdydx = ∫∫ (12−3y^2) dydx = ∫ 12(2-x)-(2-x)^3 dx =
24(2)-6(2)^2+(2-2)^4/4 -(2-0)^4/4 = 20
Step-by-step explanation:
