Question
The difference between an angle and its complement is 10° find measure of larger angle.
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Solve the following quadratic equation by factorization.
$3 x^2-2 \sqrt{6} x+2=0$
→$3 x^2-2 \sqrt{6} x+2=0$
In figure 3.98, seg AB is a diameter of a circle with centre O . The bisector of∠ ACB intersects the circle at point D. Prove that, seg AD ≅seg BD.
Complete the following proof by filling in the blanks.
Proof: Draw seg OD.
$\angle ACB =\square \ldots 90^{\circ} \ldots \ldots \ldots . . \text { angle inscribed in semicircle }$
$\angle DCB =\square \ldots . . .45^{\circ} \ldots CD \text { is the bisector of } \angle C$
$m (\operatorname{arc} DB )=\square . . . .45^{\circ} \ldots \text { inscribed angle theorem }$
$\angle DOB =\square \ldots . . . .90^{\circ} \text { definition of measure of an arc (I) }$
$\operatorname{seg} OA \cong \operatorname{seg} OB \ldots \ldots . . \text { radii of the circle... } \square \text { (II) }$
$\therefore \text { line } OD \text { is } \square \text { of seg } AB . \ldots . . \text { bisector.... From (I) and (II) }$
$\therefore \text { seg } AD \cong \operatorname{seg} BD$
→Complete the following proof by filling in the blanks.
Proof: Draw seg OD.
$\angle ACB =\square \ldots 90^{\circ} \ldots \ldots \ldots . . \text { angle inscribed in semicircle }$
$\angle DCB =\square \ldots . . .45^{\circ} \ldots CD \text { is the bisector of } \angle C$
$m (\operatorname{arc} DB )=\square . . . .45^{\circ} \ldots \text { inscribed angle theorem }$
$\angle DOB =\square \ldots . . . .90^{\circ} \text { definition of measure of an arc (I) }$
$\operatorname{seg} OA \cong \operatorname{seg} OB \ldots \ldots . . \text { radii of the circle... } \square \text { (II) }$
$\therefore \text { line } OD \text { is } \square \text { of seg } AB . \ldots . . \text { bisector.... From (I) and (II) }$
$\therefore \text { seg } AD \cong \operatorname{seg} BD$
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→The following are quadratic equations in $x$?
$\text{x}-\frac{6}{\text{x}}=3$
→$\text{x}-\frac{6}{\text{x}}=3$
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→$2y^2 + 27y + 13 = 0$
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$f(x) = 10x^4 + 17x^3 - 62x^2 + 30x - 3, g(x) = 2x^2 + 7x + 1$
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→$18^{th}$ term of the A.P. $\sqrt{2},3\sqrt{2},5\sqrt{2},\ .....$
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Verify the following:
$\sin60^\circ\cos30^\circ-\cos60^\circ\sin30^\circ=\sin30^\circ$
→Verify the following:
$\sin60^\circ\cos30^\circ-\cos60^\circ\sin30^\circ=\sin30^\circ$
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