Answer:
55
Step-by-step explanation:
since, it is a parallelogram we can find the area easily just by multiplying the base and height
A=bh
A=55m2
Answer:
1 solution
Step-by-step explanation:
4x + x + 3 = 4x (Subtract 4x from both sides)
x + 3 = 0 (Subtract 3 from both sides)
x = -3
Answer:
$2
Step-by-step explanation:
40 x 0.05 = 2
<span>96 degrees
Looking at the diagram, you have a regular pentagon on top and a regular hexagon on the bottom. Towards the right of those figures, a side is extended to create an irregularly shaped quadrilateral. And you want to fine the value of the congruent angle to the furthermost interior angle. So let's start.
Each interior angle of the pentagon has a value of 108. The supplementary angle will be 180 - 108 = 72. So one of the interior angles of the quadrilateral will be 72.
From the hexagon, each interior angle is 120 degrees. So the supplementary angle will be 180-120 = 60 degrees. That's another interior angle of the quadrilateral.
The 3rd interior angle of the quadrilateral will be 360-108-120 = 132 degrees. So we now have 3 of the interior angles which are 72, 60, and 132. Since all the interior angles will add up to 360, the 4th angle will be 360 - 72 - 60 - 132 = 96 degrees.
And since x is the opposite (or congruent) angle to this 4th interior angle, it too has the value of 96 degrees.</span>
Answer:
t = 460.52 min
Step-by-step explanation:
Here is the complete question
Consider a tank used in certain hydrodynamic experiments. After one experiment the tank contains 200 liters of a dye solution with a concentration of 1 g/liter. To prepare for the next experiment, the tank is to be rinsed with fresh water flowing in at a rate of 2 liters/min, the well-stirred solution flowing out at the same rate.Find the time that will elapse before the concentration of dye in the tank reaches 1% of its original value.
Solution
Let Q(t) represent the amount of dye at any time t. Q' represent the net rate of change of amount of dye in the tank. Q' = inflow - outflow.
inflow = 0 (since the incoming water contains no dye)
outflow = concentration × rate of water inflow
Concentration = Quantity/volume = Q/200
outflow = concentration × rate of water inflow = Q/200 g/liter × 2 liters/min = Q/100 g/min.
So, Q' = inflow - outflow = 0 - Q/100
Q' = -Q/100 This is our differential equation. We solve it as follows
Q'/Q = -1/100
∫Q'/Q = ∫-1/100
㏑Q = -t/100 + c

when t = 0, Q = 200 L × 1 g/L = 200 g

We are to find t when Q = 1% of its original value. 1% of 200 g = 0.01 × 200 = 2

㏑0.01 = -t/100
t = -100㏑0.01
t = 460.52 min