Work out the first four terms and the 10th term of the quadratic sequence with
a T(n) = 2n2 + 4
b T(n) = 3n2-3
c T(n) = -3n2 + 7
d T(n) = 4n? -5

a. The first four terms and the 10th term are 6, 12, 22, 36, , 204

b. The first four terms and the 10th term are 0, 9, 24, 45, , 297

c. The first four terms and the 10th term are 4, -5, -20, -41, , -293

d. The first four terms and the 10th term are -1, 11, 31, 59, , 395

Step-by-step explanation:

a.

∵ T(n) = 2n² + 4 is the rule of the sequence, where

∴ For the first term, n = 1

→ Substitute n by 1 in the rule above to find T(1)

∵ T(1) = 2(1)² + 4 = 2(1) + 4 = 2 + 4

∴ T(1) = 6

→ For the second term, n = 2

∵ T(2) = 2(2)² + 4 = 2(4) + 4 = 8 + 4

∴ T(2) = 12

→ For the third term, n = 3

∵ T(3) = 2(3)² + 4 = 2(9) + 4 = 18 + 4

∴ T(3) = 22

→ For the fourth term, n = 4

∵ T(4) = 2(4)² + 4 = 2(16) + 4 = 32 + 4

∴ T(4) = 36

→ For the tenth term, n = 10

∵ T(10) = 2(10)² + 4 = 2(100) + 4 = 200 + 4

∴ T(10) = 204

The first four terms and the 10th term are 6, 12, 22, 36, , 204

b.

∵ T(n) = 3n² - 3 is the rule of the sequence

→ For the first term, n = 1

∵ T(1) = 3(1)² - 3 = 3(1) - 3 = 3 - 3

∴ T(1) = 0

→ For the second term, n = 2

∵ T(2) = 3(2)² - 3 = 3(4) - 3 = 12 - 3

∴ T(2) = 9

→ For the third term, n = 3

∵ T(3) = 3(3)² - 3 = 3(9) - 3 = 27 - 3

∴ T(3) = 24

→ For the fourth term, n = 4

∵ T(4) = 3(4)² - 3 = 3(16) - 3 = 48 - 3

∴ T(4) = 45

→ For the tenth term, n = 10

∵ T(10) = 3(10)² - 3 = 3(100) - 3 = 300 - 3

∴ T(10) = 297

The first four terms and the 10th term are 0, 9, 24, 45, , 297

c.

∵ T(n) = -3n² + 7 is the rule of the sequence

→ For the first term, n = 1

∵ T(1) = -3(1)² + 7 = -3(1) + 7 = -3 + 7

∴ T(1) = 4

→ For the second term, n = 2

∵ T(2) = -3(2)² + 7 = -3(4) + 7 = -12 + 7

∴ T(2) = -5

→ For the third term, n = 3

∵ T(3) = -3(3)² + 7 = -3(9) + 7 = -27 + 7

∴ T(3) = -20

→ For the fourth term, n = 4

∵ T(4) = -3(4)² +7 = -3(16) + 7 = -48 + 7

∴ T(4) = -41

→ For the tenth term, n = 10

∵ T(10) = -3(10)² + 7 = -3(100) + 7 = -300 + 7

∴ T(10) = -293

The first four terms and the 10th term are 4, -5, -20, -41, , -293

d.

∵ T(n) = 4n² - 5 is the rule of the sequence

→ For the first term, n = 1

∵ T(1) = 4(1)² - 5 = 4(1) - 5 = 4 - 5

∴ T(1) = -1

→ For the second term, n = 2

∵ T(2) = 4(2)² - 5 = 4(4) - 5 = 16 - 5

∴ T(2) = 11

→ For the third term, n = 3

∵ T(3) = 4(3)² - 5 = 4(9) - 5 = 36 - 5

∴ T(3) = 31

→ For the fourth term, n = 4

∵ T(4) = 4(4)² - 5 = 4(16) - 5 = 64 - 5

∴ T(4) = 59

→ For the tenth term, n = 10

∵ T(10) = 4(10)² - 5 = 4(100) - 5 = 400 - 5

∴ T(10) = 395

The first four terms and the 10th term are -1, 11, 31, 59, , 395

δabc ≅ δdcb by sas because

db ≅ ac . . given (side)

bc ≅ bc . . reflexive property (side)

∠dbc ≅ ∠ acb . . alternate interior angles where transversal bc crosses parallel lines bd and ac (included angle)

greater than one

step-by-step explanation: