Calculate the deceleration (in m/s2) of a snow boarder going up a 2.65° slope assuming the coefficient of friction for waxed wood on wet snow. the equation a = g(sin(θ) − μk cos(θ)) for a snow boarder going downhill may be useful, but be careful to consider the fact that the snow boarder is going uphill. (enter the magnitude.) m/s2

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to solve this problem, we can use newton's second law

F=MxA
M is the mass of the object in motion
A is its acceleration (or deceleration)
F the sum of all exterior forces 
in our case
the exterior forces are Fk (the friction force) and P (force of gravity)

In fact, Fk is positive because the motion is upward, its value is
Fk=μk MgcosΘ
and P=MgsinΘ

so we have F=MxA= μk MgcosΘ + MgsinΘ, and from where 
A =μk gcosΘ + gsinΘ
μk=0.1 (coefficient of static friction on ice)

A=0.1x9.8xcos2.65° + 9.8xsin2.65° =0.97 + 0.45 = 1.42 m/s²

The answer is 12, for the points