Answer: 20.5
Form your equation. The formula for perimeter is (2l)(2w) or l + l + w + w, where l = length and w = width.
(l - 5) + (l - 5) + (l) + (l) = 72
Simplify and solve your equation.
4l - 10 = 72
4l = 82
l = 20.5
Since l = length, length = 20.5 and width = 15.5
You can check your work, too:
20.5 + 20.5 + 15.5 + 15.5 = 72
Plug -2 into the equation where t is, so 4 - 3( - 2 ). Multiply 3 and -2 first which is -6. 4 - (-6) is 4+6=10
Answer:4
Step-by-step explanation:
Answer:
Step-by-step explanation:
here is some helpful trig function reminders.. copy this and keep it
Use SOH CAH TOA to recall how the trig functions fit on a triangle
SOH: Sin(Ф)= Opp / Hyp
CAH: Cos(Ф)= Adj / Hyp
TOA: Tan(Ф) = Opp / Adj
they ask how long is the ladder
they give us the opposite side (8) and the adjacent side (6)
and the Hyp is the length of the ladder. They expect you to sort this out in your head. :/
there is two steps .. we'll solve for the angle.. then we can use that to find the Hyp side..(long side of triangle) which is the ladder length
Tan(Ф) = Opp / Adj
Tan(Ф) = 8 / 6
Ф = arcTan(8 / 6)
Ф ≈ 53.130°
now lets use that angle with one of the other trig functions to find the Hyp.. sin or cos will work b/c we have both the opp and the adj sides lengths. let's use cos
Cos(53.130) = 6 / Hyp
Hyp = 6 / Cos(53.130) ( I just used algebra to swap the two)
Hyp = 9.99997
the ladder will be 10' long.
ask in the comments if you need any more explaining
For three fair six-sided dice, the possible sum of the faces rolled can be any digit from 3 to 18.
For instance the minimum sum occurs when all three dices shows 1 (i.e. 1 + 1 + 1 = 3) and the maximum sum occurs when all three dces shows 6 (i.e. 6 + 6 + 6 = 18).
Thus, there are 16 possible sums when three six-sided dice are rolled.
Therefore, from the pigeonhole principle, <span>the minimum number of times you must throw three fair six-sided dice to ensure that the same sum is rolled twice is 16 + 1 = 17 times.
The pigeonhole principle states that </span><span>if n items are put into m containers, with n > m > 0, then at least one container must contain more than one item.
That is for our case, given that there are 16 possible sums when three six-sided dice is rolled, for there to be two same sums, the number of sums will be greater than 16 and the minimum number greater than 16 is 17.
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