7 to the 10 power would be 70
<span>We are given that ||e|| = 1, ||f|| = 1. </span>
<span>Since ||e + f|| = sqrt(3/2), we have </span>
<span>3/2 = (e + f) dot (e + f) </span>
<span>= (e dot e) + 2(e dot f) + (f dot f) </span>
<span>= ||e||^2 + 2(e dot f) + ||f||^2 </span>
<span>= 1^2 + 2(e dot f) + 1^2 </span>
<span>= 2 + 2(e dot f). </span>
<span>So e dot f = -1/4. </span>
<span>Therefore, </span>
<span>||2e - 3f||^2 = (2e - 3f) dot (2e - 3f) </span>
<span>= 4(e dot e) - 12(e dot f) + 9(f dot f) </span>
<span>= 4||e||^2 - 12(e dot f) + 9||f||^2 </span>
<span>= 4(1)^2 - 12(-1/4) + 9(1)^2 </span>
<span>= 4 + 3 + 9 </span>
<span>= 16. </span>
If two dices are rolled once, you would have a total number of 6*6=36 possibilities.
To get a sum greater than 10:
A) one of your dices could be showing a 5, the other a 6. That’s two possibilities. Dice A being the 5 or dice B being the 5.
B) both of your dices could be showing 6s.
That’s one possibility.
So your overall possibility to get a sum greater than 10 is (1+2)/36 3/36=1/12
One twelfth.
Answer:
neither
Step-by-step explanation:
