Answer:
131. you add them
Step-by-step explanation:
Check the picture below.
you can pretty much just count off the grid the units for JK and MI.
now, let's check how long are KI and JM
![\bf \textit{distance between 2 points}\\ \quad \\ \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) K&({{ -4}}\quad ,&{{ 4}})\quad % (c,d) I&({{ -2}}\quad ,&{{ 3}}) \end{array}\qquad % distance value d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2} \\\\\\ KI=\sqrt{[-2-(-4)]^2+[3-4]^2}\implies KI=\sqrt{(-2+4)^2+(3-4)^2} \\\\\\ KI=\sqrt{2^2+(-1)^2}\implies KI=\sqrt{4+1}\implies \boxed{KI=\sqrt{5}}\\\\ -------------------------------](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bdistance%20between%202%20points%7D%5C%5C%20%5Cquad%20%5C%5C%0A%5Cbegin%7Barray%7D%7Blllll%7D%0A%26x_1%26y_1%26x_2%26y_2%5C%5C%0A%25%20%20%28a%2Cb%29%0AK%26%28%7B%7B%20-4%7D%7D%5Cquad%20%2C%26%7B%7B%204%7D%7D%29%5Cquad%20%0A%25%20%20%28c%2Cd%29%0AI%26%28%7B%7B%20-2%7D%7D%5Cquad%20%2C%26%7B%7B%203%7D%7D%29%0A%5Cend%7Barray%7D%5Cqquad%20%0A%25%20%20distance%20value%0Ad%20%3D%20%5Csqrt%7B%28%7B%7B%20x_2%7D%7D-%7B%7B%20x_1%7D%7D%29%5E2%20%2B%20%28%7B%7B%20y_2%7D%7D-%7B%7B%20y_1%7D%7D%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0AKI%3D%5Csqrt%7B%5B-2-%28-4%29%5D%5E2%2B%5B3-4%5D%5E2%7D%5Cimplies%20KI%3D%5Csqrt%7B%28-2%2B4%29%5E2%2B%283-4%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0AKI%3D%5Csqrt%7B2%5E2%2B%28-1%29%5E2%7D%5Cimplies%20KI%3D%5Csqrt%7B4%2B1%7D%5Cimplies%20%5Cboxed%7BKI%3D%5Csqrt%7B5%7D%7D%5C%5C%5C%5C%0A-------------------------------)
![\bf \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) J&({{ -7}}\quad ,&{{ 4}})\quad % (c,d) M&({{ -8}}\quad ,&{{ 3}}) \end{array}\qquad % distance value \\\\\\ JM=\sqrt{[-8-(-7)]^2+[3-4]^2}\implies JM=\sqrt{(-8+7)^2+(3-4)^2} \\\\\\ JM=\sqrt{(-1)^2+(-1)^2}\implies \boxed{JM=\sqrt{2}}](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Barray%7D%7Blllll%7D%0A%26x_1%26y_1%26x_2%26y_2%5C%5C%0A%25%20%20%28a%2Cb%29%0AJ%26%28%7B%7B%20-7%7D%7D%5Cquad%20%2C%26%7B%7B%204%7D%7D%29%5Cquad%20%0A%25%20%20%28c%2Cd%29%0AM%26%28%7B%7B%20-8%7D%7D%5Cquad%20%2C%26%7B%7B%203%7D%7D%29%0A%5Cend%7Barray%7D%5Cqquad%20%0A%25%20%20distance%20value%0A%5C%5C%5C%5C%5C%5C%0AJM%3D%5Csqrt%7B%5B-8-%28-7%29%5D%5E2%2B%5B3-4%5D%5E2%7D%5Cimplies%20JM%3D%5Csqrt%7B%28-8%2B7%29%5E2%2B%283-4%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0AJM%3D%5Csqrt%7B%28-1%29%5E2%2B%28-1%29%5E2%7D%5Cimplies%20%5Cboxed%7BJM%3D%5Csqrt%7B2%7D%7D)
so, add all sides, and that's the perimeter of the trapezoid.
A=2πrh+2πrsquare
=2π3*5+2π3*3
30π+18π
48πmsquare
Answer:
9) 1/42
10) 1/14
Step-by-step explanation:
The probability of the compound event is the product of the probabilities of the parts. Note that the first draw (without replacement) modifies the probability of associated with the second draw.
<h3>9.</h3>
5 is one of 7 tiles. After drawing 5, 6 is one of 6 tiles.
P(5 then 6) = P(5) × P(6 | 5) = 1/7 × 1/6 = 1/42
<h3>10.</h3>
There are 3 odd tiles among the 7. After drawing one of them, 20 is one of 6 tiles.
P(odd then 20) = P(odd) × P(20 | odd) = 3/7 × 1/6 = 3/42 = 1/14
_____
Alternatively, you can consider the number of permutations of 2 tiles out of 7. That is P(7, 2) = 7!/(7-2)! = 7·6 = 42. Then the trick is to count how many of them will be the sequence of interest.
5 then 6: Among the 42 ways 2 tiles can be drawn, there is only one that is the required sequence: P(5,6) = 1/42.
odd then 20: There are 3 odd numbers, so the possible sequences of interest are (5,20), (7,20), (9,20). That is, there are 3 of 42 sequential draws that match the criteria. P(odd,20) = 3/42 = 1/14.
Is this A.P ?
Please specify mate
