Step-by-step explanation:
Transitive property of Angle Congruence
Let the length of unknown side be x
So, hypotenuse = 2x + 1
By pythagorean theorem ie , a² = b² + c²
(2x + 1)² = x² + 15²
4x² + 1 + 4x = x² + 225
3x² + 4x -224
(3x +28)(x-8)
x = -9.33 , 8
Since, length can't be negative
x = 8
∴Unknown side length = 8 m
Length of hypotenuse = 2 x 8 + 1 = 17 m
Hope it helps now.
There really isn't a right or wrong answer since this is a prediction but to solve this look at the graph and try to see the pattern
Play usually continues 7.Qf3+ Ke6 8.Nc3 (see diagram). Black will play 8...Nb4 or 8...Ne7 and follow up with c6, bolstering his pinned knight on d5. If Black plays 8...Nb4, White can force the b4 knight to abandon protection of the d5 knight with 9.a3?! Nxc2+ 10.Kd1 Nxa1 11.Nxd5, sacrificing a rook, but current analysis suggests that the alternatives 9.Qe4, 9.Bb3 and 9.O-O are stronger. White has a strong attack, but it has not been proven yet to be decisive.
Because defence is harder to play than attack in this variation when given short time limits, the Fried Liver is dangerous for Black in over-the-board play, if using a short time control. It is also especially effective against weaker players who may not be able to find the correct defences. Sometimes Black invites White to play the Fried Liver Attack in correspondence chess or in over-the-board games with longer time limits (or no time limit), as the relaxed pace affords Black a better opportunity to refute the White sacrifice.
Answer:
-13
you're basically adding -3 and -10 because they're both negatives
