What is the surface area of this square pyramid? HELP!
1 answer:
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We are given
Vertical asymptotes:
Firstly, we will factor numerator and denominator
we get
We can see that (x-3) is common in both numerator and denominator
so, we will only set x+3 to 0
and then we can find vertical asymptote
Hole:
We can see that (x-3) is common in both numerator and denominator
so, hole will be at x-3=0
Horizontal asymptote:
We can see that degree of numerator is 2
degree of denominator is also 2
for finding horizontal asymptote, we find ratio of leading coefficients of numerator and denominator
and we get
y=1
now, we can draw graph
Graph:
Answer:
(a)2(89x+84)
(b)
Step-by-step explanation:
The dimensions of the larger rectangular field are:
- Length=5x + 12; Width = 9x + 14.
The dimensions of the smaller rectangular soccer field are:
- Length=5x; Width = 9x.
(a)Area of the part of the field that is outside the soccer field
=Area of the larger rectangular field - Area of the Soccer Field
=(5x+12)(9x+14)-5x(9x)
=(5x)(9x)+70x+108x+168-5x(9x)
=178x+168
=2(89x+84)
(b)Radius of the Semicircular Fountain =2x
From Part (a),
Area of the larger rectangular field - Area of the Soccer Field=178x+168
Area of the Semicircular Fountain =
Area of the Field that does not include the soccer field or the fountain.
=Area of the larger rectangular field - Area of the Soccer Field-Area of the Semicircular Fountain