The correct answer is: $ \displaystyle x$ = 1, 2, –3 Explanation: We are asked to solve the equation x³ – 7x + 6 = 0, which is a cubic equation. We will find the roots using the trial-and-error method followed by factorization. 1. Check for a Root via Substitution: We can start by trying […]
± 4 Explanation: According to the question, the distance between the points (4, p) and (1, 0) = 5 i.e., $ \displaystyle \sqrt{{{{{(1-4)}}^{2}}+{{{(0-p)}}^{2}}}}=5$ [∵ distance between the points $ \displaystyle {{x}_{1}},{{y}_{1}}$ and $ \displaystyle {{x}_{2}},{{y}_{2}}$, d= $ \displaystyle \sqrt{{{{{\left( {{{x}_{2}}-{{x}_{1}}} \right)}}^{2}}+{{{\left( {{{y}_{2}}-{{y}_{1}}} \right)}}^{2}}}}$​] $ \displaystyle \begin{array}{l}\Rightarrow \sqrt{{{{{(-3)}}^{2}}+{{p}^{2}}}}=5\\\Rightarrow \sqrt{{9+{{p}^{2}}}}=5\end{array}$​ On squaring both the sides, we get
The correct answer is: 2 Explanation: Given, ​DE∥AB ∴ $ \displaystyle \dfrac{{\text{CD}}}{{\text{AD}}}=\dfrac{{\text{CE}}}{{\text{BE}}}$​​ [by basic proportionality theorem] ⇒ $ \displaystyle \dfrac{{x+3}}{{3x+19}}=\dfrac{x}{{3x+4}}$ ⇒(x + 3)(3x + 4) = x(3x + 19) ⇒ 3$ \displaystyle {{x}^{2}}$ + 4x + 9x + 12 = 3$ \displaystyle {{x}^{2}}$ + 19x 19x − 13x = 12 6x = 12 x = $
In △ADC and △BAC, ∠ADC = ∠BAC (Already given) ∠ACD = ∠BCA (Common angles) ∴ △ADC ∼ △BAC (AA similarity criterion) We know that corresponding sides of similar triangles are in proportion. $ \displaystyle \dfrac{{\text{CA}}}{{\text{CB}}}=\dfrac{{\text{CD}}}{{\text{CA}}}$ ​∴ $ \displaystyle {\Rightarrow \text{C}{{\text{A}}^{2}}=\text{CB}\text{.CD}\text{.}}$ Hence, proved.
The correct answer is: Sublimation Explanation: Radius of bigger circle $ \displaystyle {{{r}_{1}}}$​ = OA = 7 Radius of smaller circle $ \displaystyle {{{r}_{2}}}$ = $ \displaystyle \dfrac{{{{r}_{1}}}}{2}$ = $ \displaystyle \dfrac{7}{2}$​ Area of the bigger circle $ \displaystyle {{{C}_{1}}}$​ = $ \displaystyle {\pi r_{1}^{2}}$​ ​= $ \displaystyle \dfrac{22}{7}$ ​× 7 × 7 = 154 $ \displaystyle \text{c}{{\text{m}}^{2}}$​
The correct answer is: $ \displaystyle 3\sqrt{3}$ cm Explanation: Let P be an external point and a pair of tangents is drawn from point P and angle between these two tangents is 60°. Join OA and OP. Also, OP is a bisector line of ∠APC. ∴ ​∠APO = ∠CPO = 30°  Also, OA⊥AP​ Tangent at any