Askier leaves an 8-foot-tall ramp with an initial vertical velocity of 28 feet per second. the function h = −16t^2 28t 8 represents the height h (in feet) of the skier after t seconds. the skier has a perfect landing. how many points does the skier earn? 1 point per foot in the air, 5 points per second in the air, and a perfect landing is 25 points.

The skier earns 35.875 points.

We can find the height in the air by using -b/2a:
-28/2(-16) = -28/-32 = 0.875

This will give the skier 0.875 points.

To find the amount of time in the air, we solve the related equation:
0=-16t²+28t+8

We will first factor out the GCF, -4:
0=-4(4t²-7t-2)

Now we will factor the trinomial in parentheses using grouping.  We want factors of 4(-2)=-8 that sum to -7; -8(1) = -8 and -8+1=-7.  This is how we will "split up" bx:
0=-4(4t²-8t+1t-2)

Now we will group the first two and last two terms:
0=-4[(4t²-8t)+(1t-2)]

We will factor out the GCF of each group:
0=-4[4t(t-2)+1(t-2)]

This gives us the factored form:
0=-4(4t+1)(t-2)

Using the zero product property, we know that either t-2=0 or 4t+1=0:
t-2=0
t-2+2=0+2
t=2

4t+1=0
4t+1-1=0-1
4t=-1
4t/4 = -1/4
t=-1/4

Negative time makes no sense, so t=2.  This gives the skier 5(2) = 10 points.

Counting the perfect landing, we have 25+10+0.875 = 35.875 points.

Skew lines(segments) are lines(segments) in 3 dimensions that do not intersect the given segment and are not parallel to it.  ae, bf, eg, and fh are your skew segments.  i hope that : )