(y-14)(y+14). You just square root 196
1. Open the compass to a little more than halfway across the line segment XY. Draw an arc centered at the first endpoint X across the line segment XY. Without changing the width of the compass, place the compass tip on the
second endpoint Y. Draw a second arc across the line segment XY.
2. Line up a straightedge with the intersection of the arcs above the line XY,
and the intersection of the arcs below the line. Draw a line connecting
these two points. The line you draw is a perpendicular bisector. It
bisects the line XY at a right angle.
3. Use a compass and straightedge to construct the bisectors of the line YZ as you did with the first line segment. Extend the bisectors long enough that they intersect. The point of their intersection is the center of the circle.
4. The radius of a circle is the distance from the center to any point on the circle’s edge.
To set the width, place the tip of the compass on the center of the
circle, and open the compass to any one of your original points.Swing the compass around 360 degrees so that it draws a complete circle. The circle should pass through all three points.
Answer: The area is 12.55
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
The Volume for Rectangular prism A is 172 inches², Volume = 112 inches³
Surface Area to Volume Ratio = 43/28
<h3>Mensuration of Solids</h3>
Given Data
Rectangular prism A
Length = 2 inches
Width = 7 inches
Height = 8 inches
Surface Area
A=2(wl+hl+hw)
A=2(7*2+8*2+8*7)
A=2(14+16+56)
A= 172 inches²
Volume
Volume = L*W*H
Volume = 2*7*8
Volume = 112 inches³
Surface Area to Volume Ratio
= 172/112
= 43/28
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