<span>The production of an electric current by moving a loop of wire through a magnetic field or moving a magnet through a wire loop is called electromagnetic induction.</span>
Answer:
A half-filled container will weight 160 oz
Step-by-step explanation:
Weight of container with fruit punch = 275 oz
weight of container excluding fruit punch = 45 oz
Weight of fruit punch = Weight of container with fruit punch - weight of container excluding fruit punch = 275 oz - 45 oz = 230 oz
Thus, when container is full, it contains 230 oz.
When container will be half full, it will contain half of 230 oz
half of 230 oz = 1/2 * 230 oz = 115 oz
Thus,
a half-filled container will weight = weight of container + half of the capacity of container(115 oz) = (45 + 115) oz = 160 oz.
Thus, a half-filled container will weight 160 oz.
Answer:
The answer is 3/4
Step-by-step explanation:
Answer:
Dimensions of the container:
x = 3 m
y = 6 m
h = 1.1 m
C(min) = 270 $
Step-by-step explanation:
Volume of storage container
V = 20 m³
Let "y" be the length and "x" the width then y = 2*x
V = x*y*h ⇒ V = 2*x²*h ⇒ 20 = 2*x²+h ⇒ h = 10/ x²
Costs:
Total cost = cost of base ( 5*2*x² ) + cost of side with base x ( 2*9*x*h) +
cost of side witn base y =2x (2*9*2x*h)
C(t) = 10*x² + 18*x*h + 36*x*h
C(x) = 10x² + 54*x*10/x² ⇒ C(x) 10*x² + 540 /x
Taking derivatives on both sides of the equation we get:
C´(x) = 20*x - 540/x²
C´(x) = 0 ⇒ 20*x - 540/x² = 0 ⇒ 2x - 54/x² = 0
2x³ - 54 = 0
x³ = 27 x = 3 m
Then y = 2*x ⇒ y = 2*3 y = 6 and h = 10 / x² h = 1.1 m
And the minimum cost is
C (min) = 10*x² + 540/x ⇒ C (min) = 90 + 180
C(min) = 270 $
Answer: The answer is B, 6 1/2
Step-by-step explanation:
6+5= 11
1/2+1/2= 1
11+1= 12
12-3=9
