(t+4)/(t+5) = (t-5)/2t multiply through by 2t(t+5)
2t( t+4) = (t-5)(t+5)
2t^2 + 8t = t^2 - 25 re-arrange
t^2 + 8t + 25 = 0 Use quadratic formula to find t = -4 +- 3i
For which value of x must the expression
82x
be further simplified?
still not sure how to find the perimeter of the dodecahedron xd
Thanks!!!
Squared, so technically two dodecahedrons, I think.
Hint:
Phase Shift: −π/6 (π/6 to the left)
Vertical Shift: 0
Here is an image of the graph: https://www.mathway.com/images/placeholder.gif
THANKS MATHWAY!
Would the answer just be equal to the perimeter of the dodecahedron times two?
I could find only 3 such multiples:
60 = (1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60) > 12 (divisors)
90 = (1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90) > 12 (divisors)
150 = (1, 2, 3, 5, 6, 10, 15, 25, 30, 50, 75, 150) > 12 (divisors)
(2)
The interior angles of the pentagon each measure 108°
So.... base angles PBA and PAB are both supplemental to 108° = 72°
So angle BPA = 180 - 2(72) = 180 -144 = 36°
the PLUS pi/6 will shift the parent function to the LEFT
Nice, Guest! Another hint would be:
Since a dodecahedron is inscribed in a circle with radius one, \((P_1 P_2)^2 + (P_1 P_3)^2 + \dots + (P_{11} P_{12})^2.\) would be TWO dodecahedrons.
Thanks much!!! You guys are great.
x^2 + y^2 -4 = 0 and (x-3)^2 + (y-5)^2 - 25 = 0 equate the two and expand and simplify to get
6x + 10y-13 = 0 this is a line
10 y = -6x + 13
y = -6/10 x + 13 slope = -6/10 = -3/5 (I used desmos to look this over graphically to verify this solution)
Just as Chris found !
Hint; " Sums of squares of lengths should make you think of what?"
1) NM is the midline of the trapezoid
Its length is the average of the two base lengths....so....
(AB + DC) / 2 = 11
AB + DC = 22
5 + DC = 22
DC = 17
Thanks, jfan !!!!
Here's another way to find (3)
Call the side of the larger hexagon = S
Using the Law of Cosines we can find the side of the smaller hexagon,s, as
s^2 = [(1/2)S]^2 + [(1/2)S]^2 - 2 [ (1/2)S (1/2)S ] cos (120°)
s^2 = (1/2)S^2 -(1/2) S^2 * ( -1/2)
s^2 = (3/4)S^2
s = (√3/2) S
Because the hexagons are similar, then the scale factor of the smaller hexagon to the larger hexagon =
(√3/2) S
_______ = √3/2
S
Then the area of the smaller hexagon to the larger = square of the scale factor = 3/4
some of your posts dodnt get claimed
Thanks, CPhill!
See here :
https://web2.0calc.com/questions/help_37192
If 10% evaporates each day.....then 90% of the previous day's amount remain
So
(.90)^3 = .729 = 72.9% of the original amount remains after 3 days
We can use the tangent here....let H be the height of the cloud
tan (37° 55') = H / 1100 multiply both sides by 1100
1100 * tan (37° 55') = 1100 * tan ( 37 + 11/12)° = H ≈ 856.8 ft
Hey....I miss it, too !!!!!
Thanks for the thanks !!!!!!
This is unreadable. Please repost your question with proper formatting so it's readable
(5)
Central angle BOC = 360° - 106° - 124° = 130°
Since angle BAC is an inscribed angle intercepting the same arc as central angle BOC, it has (1/2) the measure of angle BOC = (1/2) (130°) = 65°
(4)
Angle ACD = (1/2) minor arc AC
Note that angle ABC is an inscribed angle.....so......the arc that it intercepts ( minor arc AC) = twice this = 114°
Angle ACD = (1/2) minor arc AC = (1/2)(114°) = 57°
