In both problems, the sum of side lengths is the perimeter. Opposite sides of a parallelogram (or rectangle) are equal in length, so you can find the perimeter by doubling the sum of adjacent sides.
25. 2(x +(x +15)) = (x +45) +(x +40) +(x +25)
.. 4x +30 = 3x +110 . . . . . . . . . . . . . . . . . . . . . . simplify
.. x = 80 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . subtract 3x+30
.. 4x +30 = 4*80 +30 = 240
The perimeter of each is 240 units.
26. 2(x +(x +2)) = (x) +(x +6) +(x +4)
.. 4x +4 = 3x +10 . . . . . . . . . . . . . . . . . . . . . . simplify
.. x = 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . subtract 3x+4
.. 4x +4 = 4*6 +4 = 28
The perimeter of each is 28 units.
<span>N(t) = 16t ; Distance north of spot at time t for the liner.
W(t) = 14(t-1); Distance west of spot at time t for the tanker.
d(t) = sqrt(N(t)^2 + W(t)^2) ; Distance between both ships at time t.
Let's create a function to express the distance north of the spot that the luxury liner is at time t. We will use the value t as representing "the number of hours since 2 p.m." Since the liner was there at exactly 2 p.m. and is traveling 16 kph, the function is
N(t) = 16t
Now let's create the same function for how far west the tanker is from the spot. Since the tanker was there at 3 p.m. (t = 1 by the definition above), the function is slightly more complicated, and is
W(t) = 14(t-1)
The distance between the 2 ships is easy. Just use the pythagorean theorem. So
d(t) = sqrt(N(t)^2 + W(t)^2)
If you want the function for d() to be expanded, just substitute the other functions, so
d(t) = sqrt((16t)^2 + (14(t-1))^2)
d(t) = sqrt(256t^2 + (14t-14)^2)
d(t) = sqrt(256t^2 + (196t^2 - 392t + 196) )
d(t) = sqrt(452t^2 - 392t + 196)</span>
Good evening ,
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Answer:
a=3
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Step-by-step explanation:
Look at the photo below for the answer.
:)
-4g is what I got if ur evaluating
If it’s something about slope I got -4
