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:
perimeter = 20 m
Step-by-step explanation:
given the area of the square = 25 m² and
area of a square = s² ( where s is the side length ), then
s² = 25 ( take the square root of both sides )
s = 
= 5 ← length of side
perimeter = 4s = 4 × 5 = 20 m
Answer:
I'm pretty sure it's c but I'm not sure
To solve we have the following equation below,
Assumption: Line of reference is the surface of the water.
Final distance between Jana & fishing line = (dock height) +(reels out) -(reels back)
Final distance between Jana & fishing line = 10 + 35+ 29 - 7 = 67 feet
<em> ANSWER: 67 feet</em>
Slope-intercept form is y = mx + b, where m is the slope and b is the y-intercept. We are given the slope of -3, but we need the y-intercept. We will sub in the x coordinate for x in the equation and the y coordinate for y in the equation, along with the slope and solve for b. -2 = -3(0) + b and -2 = b. Rewriting the equation, we have y = -3x - 2.
