<span><span><span>2<span>c5</span></span>+<span>44<span>c4</span></span></span>+<span>242<span>c3</span></span></span><span><span><span>2<span>c5</span></span>+<span>44<span>c4</span></span></span>+<span>242<span>c3</span></span></span><span>=<span><span><span>2<span>c3</span></span><span>(<span>c+11</span>)</span></span><span>(<span>c+11</span><span>)</span></span></span></span>
The sample space has 36 possible pairs from 1,1 1,2 1,3 up to 6,5 and 6,6
(a). Three pairs add to 4 1,3 2,2 and 3,1 so P(4) = 3/36 = 1/12
(b). 6 pairs add to 7 so P(7) = 6/36 = 1/6
(c) 15 pairs add to less than 7 so P(<7) = 15/36 = 5/12
<u>Answer</u>:- No.
<u>Explanation</u> :-
<u>Substitute these numbers in pythagoras theorem to check if the set of numbers is a pythagorean triplet.</u>
<u>Pythagoras theorem</u> :- sq. of hypotenuse (longest side) is equal to the sum of sq.s of other two sides.
<u>Here</u>,
hypotenuse = 12 (as it is the longest side)
and other two sides are 6 and 9.
----> 6^2 + 9^2 = 12^2
----> 36 + 81 = 144
----> 117 = 144
Since, LHS is not equal to RHS, this set of numbers is not a pythagorean triplet.
Answer:
x + 17 = 25
Step-by-step explanation:
im pretty sure you can substitute x with any other letter but ldk
