The answer to the quest is 8.09
Answer:
B, C, and D are all equivalent to 3(4h+2k)
Step-by-step explanation:
We use the distributive property to say 3(4h+2k)=12h+6k
"A" distributes to 12k+6h
"B" distributes to 12h+6k
"C" distributes to 6k+12h (associative property works)
"D" is already distributed to 12h+6k
So A is wrong, and B, C, and D are right
Hope this made sense!
Option B:
The perimeter of ΔABC is 28 units.
Solution:
AD = 5, DC = 6 and AB = 8
AD and AE are tangents to a circle from an external point A.
BE and BF are tangents to a circle from an external point B.
CD and CF are tangents to a circle from an external point C.
<em>Tangents drawn from an external point to a circle are equal in length.</em>
⇒ AD = AE, BE = BF and CD = CF
AE = 5
AE + BE = AB
5 + BE = 8
Subtract 5 from both sides.
BE = 3
BE = BF
⇒ BF = 3
CD = CF
⇒ CF = 6
Perimeter of the polygon = AE + BE + BF + CF + CD + AD
= 5 + 3 + 3 + 6 + 6 + 5
= 28
The perimeter of ΔABC is 28 units.
Option B is the correct answer.
m∠RSV = 3x + 5, m∠RST = 8x - 14 <em>given</em>
m∠RSV ≅ m∠VST <em>definition of segment bisector</em>
m∠RSV + m∠VST = m∠RST <em>angle addition postulate</em>
(3x + 5) + (3x + 5) = (8x - 14) <em>substitution</em>
6x + 10 = 8x - 14 <em>simplify (added like terms)</em>
10 = 2x - 14 <em>subtraction property of equality</em>
24 = 2x <em>addition property of equality</em>
12 = x <em>division property of equality</em>
Answer: x = 12
Answer:
about 17 meters
Step-by-step explanation:
We can use the Pythagorean theorem to put an upper bound on the height of the bump in the rail. This assumes half the expanded rail length (d+e) is the hypotenuse of a right triangle whose legs are the bump height (b) and the 2500 meter distance (d) from the center of the rail to its end.
The Pythagorean theorem relates these distances this way:
b^2 + d^2 = (d+e)^2
Expanding the square on the right, we can simplify the expression to find b.
b^2 = (d^2 +2de +e^2) -d^2
b^2 = e(2d +e)
b = √(e(2d +e))
Using lengths in meters, we can fill this in to calculate b.
b = √(.06(2·2500 +.06)) = √300.0036
b ≈ 17.32 . . . . meters
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<em>Comment on this solution</em>
We don't expect rails to tear loose from the rail bed and rise up to a height matching that of a 3-story building. That is why there are typically expansion joints and shorter rail lengths used in the construction of railways.
The height is a little lower if we take physics into account and distribute the stress in the rail along its length. No doubt the final curve is somewhat more complicated than the triangle we have assumed.
If it were an ellipse, the height might only be 9.4 meters, with the steepest rise occurring near the ends of the rail. The math for this model is beyond the scope of this answer.
