Answer:
I believe this is neither a direct or inverse relationship.
Step-by-step explanation:
Answer: 
Step-by-step explanation:
To polygons are said to be similar if all the corresponding angles of given polygons are equal.
In the given figure, it can be seen that in quadrilateral FSHB and quadrilateral KTWJ all the corresponding angles are equal.
∠F=∠K
∠S=∠T
∠H=∠W
∠B=∠J
Therefore, The given polygons are similar.
Hence,
A bisector is a line that divides either a line or an angle into <em>two</em><em> proportionate</em> parts or angles. Thus, Anton's <em>bisector</em> would divide the segment into two <u>equal parts</u>, while Maxim's <em>bisector</em> would divide the angle into two <u>equal angles</u>.
The <u>similarities</u> between their construction are:
- <em>Intersecting</em> arcs through which the bisector would pass are required.
- The arcs are dawn using <em>the same</em> radius of any measure.
- The <em>edges</em> of the arc of the given angle, and the ends of the segment are used as <em>centers</em>.
The <u>differences</u> between their construction are:
- Anton has to draw two intersecting arcs <u>above</u> and <u>below</u> the segment. While Maxim would draw two intersecting arcs <u>within</u> the lines forming the angles.
- Anton's bisector would be <em>perpendicular</em> to the segment, while Maxim's bisector would be at <em>an angle</em> which is half of the initial angle.
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A= 30
Explanation: 88-58=30
Let S(t) denote the amount of sugar in the tank at time t. Sugar flows in at a rate of
(0.04 kg/L) * (2 L/min) = 0.08 kg/min = 8/100 kg/min
and flows out at a rate of
(S(t)/1600 kg/L) * (2 L/min) = S(t)/800 kg/min
Then the net flow rate is governed by the differential equation
Solve for S(t):
The left side is the derivative of a product:
![\dfrac{\mathrm d}{\mathrm dt}\left[e^{t/800}S(t)\right]=\dfrac8{100}e^{t/800}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dt%7D%5Cleft%5Be%5E%7Bt%2F800%7DS%28t%29%5Cright%5D%3D%5Cdfrac8%7B100%7De%5E%7Bt%2F800%7D)
Integrate both sides:
There's no sugar in the water at the start, so (a) S(0) = 0, which gives
and so (b) the amount of sugar in the tank at time t is
As 
, the exponential term vanishes and (c) the tank will eventually contain 64 kg of sugar.
