Differentiate tan x2 with respect to x.
The given function is \(\begin{array}{l}\begin{array}{l} y=\tan x^{2}\\ \text { Differentiation w.r.t. } \mathbf{x}\\ \frac{d y}{d x}=\sec ^{2}\left(x^{2}\right) \times 2 x\\... View Article
The given function is \(\begin{array}{l}\begin{array}{l} y=\tan x^{2}\\ \text { Differentiation w.r.t. } \mathbf{x}\\ \frac{d y}{d x}=\sec ^{2}\left(x^{2}\right) \times 2 x\\... View Article
The given expression can be rewritten as \(\begin{array}{l}\mathbf{a}^{3}+(-\mathbf{b})^{3}+\mathbf{c}^{3}-\mathbf{3 a}(-\mathbf{b}) \mathbf{c}\\ \text { This in the familiar form, where b is... View Article
1 cm = 10 mm
The roots of the quadratic equation are found as follows. \(\begin{array}{l}\begin{array}{l} \Rightarrow \boldsymbol{\Delta }=\mathbf{b}^{2}-\mathbf{4 a c}=\mathbf{0} \\ \Rightarrow(\mathbf{k}+\mathbf{1})^{2}-4(\mathbf{k}+\mathbf{4}) \times \mathbf{1}=\mathbf{0}... View Article
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The given statement is true. The diagonals of a square are perpendicular.
The highest power of x is 1, hence the degree of the polynomial 2x – 1 is 1.
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