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[tex] \sf \large \: Polynomials[/tex]
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
[tex]\large\underline\mathfrak\green{Polynomials}[/tex]
Option b), 1
[tex]\rule{190}1[/tex]
Answer 2 :
Option d), -1
[tex]\rule{190}1[/tex]
Answer 3 :
[tex]\implies[/tex]Option a), π/2
[tex]\rule{190}1[/tex]
Answer 4 :
[tex]\implies[/tex]Option b), 1
[tex]\rule{190}1[/tex]
Answer 5 :
[tex]\implies[/tex]Option c), 2
[tex]\rule{190}1[/tex]
Answer 6 :
[tex]\implies[/tex]Option a), 1
[tex]\rule{190}1[/tex]
Answer 7 :
[tex]\implies[/tex]Option a), 3
Explanation :
p(0) = 5x - 4x² + 3
[tex]\implies[/tex]5(0) - 4(0)² + 3
[tex]\implies[/tex]0 - 0 + 3
[tex]\implies[/tex]3
[tex]\rule{190}1[/tex]
Answer 8 :
[tex]\implies[/tex]Option b), -6
Explanation :
p(-1) = 5x - 4x² + 3
[tex]\implies[/tex]5(-1) - 4(-1)² + 3
[tex]\implies[/tex]-5 -4(1) + 3
[tex]\implies[/tex]- 5 - 4 + 3
[tex]\implies[/tex] -9 + 3
[tex]\implies[/tex]- 6
[tex]\rule{190}1[/tex]
Answer 9 :
Option b), 0
Explanation :
[tex]\implies[/tex]x(x + 1) -1(x + 1)
[tex]\implies[/tex] x² + x - x - 1
[tex]\implies[/tex] x² - 1
p(1) = x² - 1
[tex]\implies[/tex](1)² - 1
[tex]\implies[/tex]1 - 1
[tex]\implies[/tex]0
[tex]\rule{190}1[/tex]
Answer 10 :
Option b), 2
Explanation :
p(0) = -t³ + 2t² + 0 + 2
[tex]\implies[/tex]-(0)³ + 2(-0)² + 0 + 2
[tex]\implies[/tex] -0 + 0 + 0 + 2
[tex]\implies[/tex]2
[tex]\rule{190}1[/tex]
Answer 11 :
Option a), 4
Explanation :
p(2) = -t³ + 2t² + t + 2
[tex]\implies[/tex] - (2)³ + 2(2)² + 2 + 2
[tex]\implies[/tex]- 8 + 2(4) + 2 + 2
[tex]\implies[/tex]-8 + 8 + 2 + 2
[tex]\implies[/tex] 4
[tex]\rule{190}1[/tex]
Answer 12 :
Option d), 1
Explanation :
y(0) = y² - y + 1
[tex]\implies[/tex](0)² - 0 + 1
[tex]\implies[/tex]0 - 0 + 1
[tex]\implies[/tex]1