Answer:
dy/dt = - (1/5) ft/s = - 0.2 ft/s
Step-by-step explanation:
Given
L = 5 ft
Qin = 25 ft³/s
Qout = 30 ft³/s
h = 10 ft
dy/dt = ?
We can apply the relation
ΔQ = Qint - Qout = 25 ft³/s - 30 ft³/s
⇒ ΔQ = - 5 ft³/s
Then we use the formula
Q = v*A
where Q = ΔQ, A = L² is the area of square base and v = dy/dt is the rate of change in the depth of the solution in the tank
⇒ ΔQ = (dy/dt)*L²
⇒ dy/dt = ΔQ/L²
⇒ dy/dt = (- 5 ft³/s)/(5 ft)²
⇒ dy/dt = - (1/5) ft/s = - 0.2 ft/s
7% of $9,200 is $644. Add $800 + $644 to get $1,444
ANSWER:
$1,444
Not sure the right equation so i did both....
#1 - √12m√15m=1m√2
Simplifies to:
13.416408m2=1.414214m
Let's solve your equation step-by-step.
13.416408m2=1.414214m
Step 1: Subtract 1.414214m from both sides.
13.416408m2−1.414214m=1.414214m−1.414214m
13.416408m2−1.414214m=0
Step 2: Factor left side of equation.
m(13.416408m−1.414214)=0
Step 3: Set factors equal to 0.
m=0 or 13.416408m−1.414214=0
m=0 or m=0.105409
Answer: m=0 or m=0.105409
#2 - √12m√15m
Simplifies to:
13.416408m2
=13.416408*m2
=13.416408*(m*m)
=13.416408m2
Can I get a brainliest Please & Thank you...
Answer:
Step-by-step explanation:
Given: Dimensions of a big locker are 0.5 m × 0.6 m × 1.2 m
Dimensions of a small locker are 0.5 m × 0.6 m × 
(as height of small locker is half the height of big locker )
To find: total volume of one big locker and one small locker
Solution:
Volume of cuboid = length × breadth × height
Total volume of one big locker and one small locker = Total volume of one big locker + total volume of one small locker
= 
