Math, asked by anjalisingh2006, 10 months ago

Give me answer of all question please..............

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Answers

Answered by StarrySoul
11

 \sf \large \:  Polynomials

Answer 1 :

→ Option b), 1

Explanation :

Coefficient is a numerical or constant quantity in a term,Each term of a polynomial has a coefficient.

Answer 2 :

→ Option d), -1

Explanation :

Coefficient is a numerical or constant quantity in a term,Each term of a polynomial has a coefficient.

Answer 3 :

→ Option a), π/2

Explanation :

Coefficient is a numerical or constant quantity in a term,Each term of a polynomial has a coefficient.

Answer 4 :

→ Option b), 1

Explanation :

The highest power of the variable in a polynomial is called as the degree of the polynomial.

Answer 5 :

→ Option c), 2

Explanation :

The highest power of the variable in a polynomial is called as the degree of the polynomial.

Answer 6 :

→ Option a), 1

Explanation :

The highest power of the variable in a polynomial is called as the degree of the polynomial. The degree of a non-zero constant polynomial is zero.

Answer 7 :

→ Option a), 3

Explanation :

p(0) = 5x - 4x² + 3

→ 5(0) - 4(0)² + 3

→ 0 - 0 + 3

3

Answer 8 :

→ Option b), -6

Explanation :

p(-1) = 5x - 4x² + 3

→ 5(-1) - 4(-1)² + 3

→ -5 -4(1) + 3

→ - 5 - 4 + 3

→ -9 + 3

- 6

Answer 9 :

Option b), 0

Explanation :

→ x(x + 1) -1(x + 1)

→ x² + x - x - 1

→ x² - 1

p(1) = x² - 1

→ (1)² - 1

→ 1 - 1

0

Answer 10 :

Option b), 2

Explanation :

p(0) = -t³ + 2t² + 0 + 2

→ -(0)³ + 2(-0)² + 0 + 2

→ -0 + 0 + 0 + 2

2

Answer 11 :

Option a), 4

Explanation :

p(2) = -t³ + 2t² + t + 2

→ - (2)³ + 2(2)² + 2 + 2

→ - 8 + 2(4) + 2 + 2

→ -8 + 8 + 2 + 2

4

Answer 12 :

Option d), 1

Explanation :

y(0) = y² - y + 1

→ (0)² - 0 + 1

→ 0 - 0 + 1

1


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Answered by Anonymous
3

\large\underline\mathfrak\green{Polynomials}

Option b), 1

\rule{190}1

Answer 2 :

Option d), -1

\rule{190}1

Answer 3 :

\impliesOption a), π/2

\rule{190}1

Answer 4 :

\impliesOption b), 1

\rule{190}1

Answer 5 :

\impliesOption c), 2

\rule{190}1

Answer 6 :

\impliesOption a), 1

\rule{190}1

Answer 7 :

\impliesOption a), 3

Explanation :

p(0) = 5x - 4x² + 3

\implies5(0) - 4(0)² + 3

\implies0 - 0 + 3

\implies3

\rule{190}1

Answer 8 :

\impliesOption b), -6

Explanation :

p(-1) = 5x - 4x² + 3

\implies5(-1) - 4(-1)² + 3

\implies-5 -4(1) + 3

\implies- 5 - 4 + 3

\implies -9 + 3

\implies- 6

\rule{190}1

Answer 9 :

Option b), 0

Explanation :

\impliesx(x + 1) -1(x + 1)

\implies x² + x - x - 1

\implies x² - 1

p(1) = x² - 1

\implies(1)² - 1

\implies1 - 1

\implies0

\rule{190}1

Answer 10 :

Option b), 2

Explanation :

p(0) = -t³ + 2t² + 0 + 2

\implies-(0)³ + 2(-0)² + 0 + 2

\implies -0 + 0 + 0 + 2

\implies2

\rule{190}1

Answer 11 :

Option a), 4

Explanation :

p(2) = -t³ + 2t² + t + 2

\implies - (2)³ + 2(2)² + 2 + 2

\implies- 8 + 2(4) + 2 + 2

\implies-8 + 8 + 2 + 2

\implies 4

\rule{190}1

Answer 12 :

Option d), 1

Explanation :

y(0) = y² - y + 1

\implies(0)² - 0 + 1

\implies0 - 0 + 1

\implies1

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