Question

find the values of P (1/2),p(-3/5), p(-1.5) p(0.02 ) p(0) in p(n)=2n2-3n+1​

Answer 1

Answer:

ANSWER

For collinearity of points A(−1,3),B(2,p),C(5,−1), area of ΔABC should be zero ,i.,e,

2

1

∣(p−1)+2(−1−3)+5(3−p)∣=0

⇒p−1−8+15−5p=0

⇒−6p+6=0

⇒−6p=−6

⇒p=1

Answer 2

Step-by-step explanation:

$given : p(n) = 2n {}^{2} - 3n + 1$

$1) \: p(1 \div 2) = 2 \times (1 \div 2) {}^{2} - 3 \times (1 \div 2) + 1$

$= 2 \times (1 \div 4) - 3 \div 2 + 1$

$1 \div 2 - 3 \div 2 + 1$

$(1 - 3) \div 2 + 1$

$= - 2 \div 2 + 1$

$= - 1 + 1$

$= 0$

$2) \: p( - 3 \div 5) \: = \: 2 \times ( - 3 \div 5) {}^{2} - 3 \times 3 \div 5 \: + 1$

$= ( - 6 \div 5) - 9 \div 5 + 1$

$= ( - 6 - 9) \div 5 + 1$

$= - 15 \div 5 + 1$

$= - 3 + 1$

$= - 2$

$3) \: p \: (1.5) = 2 \times (1.5) {}^{2} - 3 \times 1.5 + 1$

$= 2 \times 2.25 - 4.5 + 1$

$= 4.50 - 4.5 + 1$

$= 0 + 1$

$= 1$

$4) \: p(0.02) = 2 \times (0.02) {}^{2} - 3 \times 0.02 + 1$

$= 2 \times 0.0004 - 0.06 + 1$

$= 0.0008 - 0.06 + 1$

$= - 0.0592 + 1$

$= 0.9408$

$5) \: p(0) = 2 \times (0) {}^{2} - 3 \times 0 + 1$

$= 2 \times 0 - 3 \times 0 + 1$

$= 0 - 0 + 1$

$= 1$