Answer:
![36\sqrt{2}+30\sqrt{3}+10\sqrt{14}+8\sqrt{21}](https://tex.z-dn.net/?f=36%5Csqrt%7B2%7D%2B30%5Csqrt%7B3%7D%2B10%5Csqrt%7B14%7D%2B8%5Csqrt%7B21%7D)
Step-by-step explanation:
Use the distributive property and the rules of roots.
![(2\sqrt{7}+3\sqrt{6})(5\sqrt{2}+4\sqrt{3}) = 10\sqrt{14}+8\sqrt{21}+15\sqrt{12}+12\sqrt{18}\\\\=\boxed{36\sqrt{2}+30\sqrt{3}+10\sqrt{14}+8\sqrt{21}}](https://tex.z-dn.net/?f=%282%5Csqrt%7B7%7D%2B3%5Csqrt%7B6%7D%29%285%5Csqrt%7B2%7D%2B4%5Csqrt%7B3%7D%29%20%3D%2010%5Csqrt%7B14%7D%2B8%5Csqrt%7B21%7D%2B15%5Csqrt%7B12%7D%2B12%5Csqrt%7B18%7D%5C%5C%5C%5C%3D%5Cboxed%7B36%5Csqrt%7B2%7D%2B30%5Csqrt%7B3%7D%2B10%5Csqrt%7B14%7D%2B8%5Csqrt%7B21%7D%7D)
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(a√b)(c√d) = (ac)√(bd)
√(a²b) = a√b
Answer:the answer to this in a fraction form is 127 over 15. in decimal form is 8.466666667
In this question, the condition asked is two which was
1. The dice sum >7 (notice that >7 mean at least 8)
2. One of the two dice show a 2
It will be easier to fulfill the No.2 condition first so you can divide the probability into two,
Case 1: 1st dice show 2: then the second dice need to be 6 to made it sum>7
Case 2:2nd dice show 2: then the second dice need to be 6 to made it sum>7
The probability for a dice showing 2 is 1/6. The probability for a dice showing 6 is 1/6. Then the probability is 1/6 * 1/6= 1/36 for each case
Since case 1 and case 2 has same probability, then you just need to multiply 1/36 *2 = 1/18