Here are the first five terms of a quadratic sequence. 10 21 38 61 90 Find an expression, in terms of n, for the nth term of this sequence.

  an = 3n^2 +2n +5

Step-by-step explanation:

If the n-th term is given by ...

  an = a·n^2 +bn +c

Then we can write 3 equations in a, b, c:

  10 = a·1 +b·1 + c

  21 = a·4 +b·2 +c

  38 = a·9 +b·3 +c

Subtracting each equation from the next gives ...

  11 = 3a +b

  17 = 5a +b

Then subtracting the first of these from the second, we have ...

  6 = 2a

  3 = a

Substituting in to the first of the reduced equations, we find ...

  11 = 3(3) +b   ⇒   b = 2

And substituting these values into the first equation gives ...

  10 = 3 +2 +c   ⇒   c = 5

The expression for the n-th term is ...

  an = 3n^2 +2n +5

9) i can't see the angle so whatever it is just do this: cos(angle) x hypotenuse and then use that answer and they hypotenuse and plug it into this: (answer you just got)^2 + b^2 = (hypotenuse)^2. the first answer will be the answer of the adjacent side and the second answer after this equation will be the opposite side.

10) y = 14 x = √147 or 12.12435565

11) x = 4.8

12) x = 4.2

13) x = 9.3

14) pq: 13.2

qr : 17.6

angle p: 53º

15) angle c: 54.3º

angle a: 35.7º

16) h = 21040

v = 1841

step-by-step explanation:

10) 7/sin(30) = 14 = y

7^2 + b^2 = 14^2

49 + b^2 = 196

b^2 = 196 - 49

√b^2 = √147

x = √147

11) cos(70) x 14 = 4.788282007

x = 4.8

12) tan(25) x 9 = 4.196768923

x = 4.2

13) sin(31) x 18 = 9.270685348

x = 9.3

14) pq: sin(37) x 22 = 13.23993051

pq: 13.2

qr: cos(37) x 22 = 17.56998122

qr : 17.6

angle p: 180 - (37 + 90) = 53º

15) cos-1(7/12) = 54.31466529

angle c: 54.3º

angle a: 180 - (90 + 54.31466529) = 35.68533471

angle a: 35.7º

16) 4 x 5280 = 21120

cos(5) x 21120

h = 21039.63202

h = 21040

sin(5) x 21120

v = 1840.729287

v = 1841