9.
By the Segment Addition Postulate, SAP, we have
XY + YZ = XZ
so
YZ = XZ - XY = 5 cm - 2 cm = 3 cm
10.
M is the midpoint of XZ=5 cm so
XM = 5 cm / 2 = 2.5 cm
11.
XY + YM = XM
YM = XM - XY = 2.5 cm - 2 cm = 0.5 cm
12.
The midpoint is just the average of the coordinate A(-3,2), B(5,-4)

Answer: M is (1,-1)
You'll have to plot it yourself.
13.
For distances we calculate hypotenuses of a right triangle using the distnace formula or the Pythagorean Theorem.

Answer: AB=10
M is the midpoint of AB so
Answer: AM=MB=5
14.
B is the midpoint of AC. We have A(-3,2), B(5,-4)
B = (A+C)/2
2B = A + C
C = 2B - A
C = ( 2(5) - -3, 2(-4) - 2 ) = (13, -10)
Check the midpoint of AC:
(A+C)/2 = ( (-3 + 13)/2, (2 + -10)/2 ) = (5, -4) = B, good
Answer: C is (13, -10)
Again I'll leave the plotting to you.
Answer:
130, exterior, 8, 93, 87, interior
Step-by-step explanation:
Answer:
100 because it is corresponding angels ...
Step-by-step explanation:
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.