Write the degree of the following polynomial: $\sqrt{11} t+14$.
(1) The given polynomial is $p(t) = \sqrt{11} t + 14$.
In a polynomial,the degree is defined as the highest power of the variable present in the expression.
Here,the variable is $t$,and its exponent is $1$.
Therefore,the degree of the polynomial $\sqrt{11} t + 14$ is $1$.
In a polynomial,the degree is defined as the highest power of the variable present in the expression.
Here,the variable is $t$,and its exponent is $1$.
Therefore,the degree of the polynomial $\sqrt{11} t + 14$ is $1$.
Explore More
Similar Questions
Factorise $x^{3}+12 x^{2}+39 x+28$.
Medium
View SolutionFind the zero of the polynomial in the following case:
$p(x) = \frac{2}{3}x + \frac{5}{4}$
$p(x) = \frac{2}{3}x + \frac{5}{4}$
Easy
View SolutionFrom the following polynomials,find out which of them has $(x-1)$ as a factor:
$p(x) = 2x^3 + 5x^2 - x - 6$
$p(x) = 2x^3 + 5x^2 - x - 6$
Medium
View SolutionExpand $\left(\frac{2x}{3} + \frac{3y}{4}\right)^{2}$.
Easy
View SolutionWrite whether the following statement is True or False. Justify your answer.
$\frac{1}{\sqrt{5}} x^{\frac{1}{2}} + 1$ is a polynomial.
$\frac{1}{\sqrt{5}} x^{\frac{1}{2}} + 1$ is a polynomial.
Easy
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo