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.
Prove we are to prove 4(coshx)^3 - 3(coshx) we are asked to prove 4(coshx)^3 - 3(coshx) to be equal to cosh 3x
= 4(e^x+e^(-x))^3/8 - 3(e^x+e^(-x))/2 = e^3x /2 +3e^x /2 + 3e^(-x) /2 + e^(-3x) /2 - 3(e^x+e^(-x))/2 = e^(3x) /2 + e^(-3x) /2 = cosh(3x) = LHS Since y = cosh x satisfies the equation if we replace the "2" with cosh3x, we require cosh 3x = 2 for the solution to work.
i.e. e^(3x)/2 + e^(-3x)/2 = 2
Setting e^(3x) = u, we have u^2 + 1 - 4u = 0
u = (4 + sqrt(12)) / 2 = 2 + sqrt(3), so x = ln((2+sqrt(3))/2) /3, Or u = (4 - sqrt(12)) / 2 = 2 - sqrt(3), so x = ln((2-sqrt(3))/2) /3,
Therefore, y = cosh x = e^(ln((2+sqrt(3))/2) /3) /2 + e^(-ln((2+sqrt(3))/2) /3) /2 = (2+sqrt(3))^(1/3) / 2 + (-2-sqrt(3))^(1/3) to be equ
= 4(e^x+e^(-x))^3/8 - 3(e^x+e^(-x))/2
= e^3x /2 +3e^x /2 + 3e^(-x) /2 + e^(-3x) /2 - 3(e^x+e^(-x))/2
= e^(3x) /2 + e^(-3x) /2
= cosh(3x)
= LHS
<span>Therefore, because y = cosh x satisfies the equation IF we replace the "2" with cosh3x, we require cosh 3x = 2 for the solution to work. </span>
i.e. e^(3x)/2 + e^(-3x)/2 = 2
Setting e^(3x) = u, we have u^2 + 1 - 4u = 0
u = (4 + sqrt(12)) / 2 = 2 + sqrt(3), so x = ln((2+sqrt(3))/2) /3,
Or u = (4 - sqrt(12)) / 2 = 2 - sqrt(3), so x = ln((2-sqrt(3))/2) /3,
Therefore, y = cosh x = e^(ln((2+sqrt(3))/2) /3) /2 + e^(-ln((2+sqrt(3))/2) /3) /2
= (2+sqrt(3))^(1/3) / 2 + (-2-sqrt(3))^(1/3)
sry I just wanted the points I'm in middle school so I don't know this stuff either but can you give free brainlyest I'm soo close to my next rank I'd really appreciate it if you would
I hope this helps you
this triangle equilateral triangle
all interior angles same
all sides same
A+A+A=180
3A=180
A=60 degree same
60-60-60
