(1 point) On a piece of paper, sketch each of the following surfaces: (i) z = x2 +y2 +6 (ii) z = 3x2 Use your graphs to fill in the following descriptions of cross-sections of the surfaces. (a) For (i) (z = x2 2 + 6): Cross sections with x fixed give a downward opening parabola in a plane parallel to the yz-plane a downward opening parabola in a plane parallel to the xz-plane a downward opening parabola in a plane parallel to the xy-plane Cross sections with y fixed give Cross sections with z fixed give (b) For (ii) (z = 3x2): Cross sections with x fixed give Cross sections with y fixed give Cross sections with z fixed give an empty set, or one or two vertical lines in a plane parallel to the yz-plane> an empty set, or one or two vertical lines in a plane parallel to the xz-plane a downward opening parabola in a plane parallel to the xy-plane

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Step-by-step explanation:

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step-by-step explanation: so we look at the formula un = a + (n-1)d

un is the amount of seats

n is the nth term aka row number

a is the original sequence/row number

d is the difference between each term

so in this a=22 d=6 and n can remain as n because we are simply formjng a rule

lets put these in the equation

un=22+(n-1)(6)

this is your answer to part a

however

lets double check that this is correct, we'll try putting it into the second term

un=a+(n-1)d

u2=22+(2-1)(6)

u2=22+6

u2=28

as the question said the second row has 28 seats so now we know the formula we made is correct.

as for part b

un=a+(n-1)

its asking us to find a row with 100 seats.

remember un was the amount of seata so all we do is substitute this in.

un=100 a=22 d=6

un=a+(n-1)d

100=22+(n-1)(6)

so now we simplify to find n which is the row number.

100-22=22+(n-1)(6)-22

78=(n-1)(6)

78/6=(n-1)(6)/6

13=n-1

13+1=n-1+1

14=n

so now we know that in the 14th row there is 100 seats.

hope this and that you understood! its my first time using the app or anyone out on here : p its also me to revise so a bunch! ask more questions if you dont understand ☺