Let's solve your equation step-by-step.<span><span><span><span>2<span>x2</span></span>−<span>3x</span></span>−4</span>=0</span>Step 1: Use quadratic formula with a=2, b=-3, c=-4.<span>x=<span><span><span>−b</span>±<span>√<span><span>b2</span>−<span><span>4a</span>c</span></span></span></span><span>2a</span></span></span><span>x=<span><span><span>−<span>(<span>−3</span>)</span></span>±<span>√<span><span><span>(<span>−3</span>)</span>2</span>−<span><span>4<span>(2)</span></span><span>(<span>−4</span>)</span></span></span></span></span><span>2<span>(2)</span></span></span></span><span>x=<span><span>3±<span>√41</span></span>4</span></span><span><span>x=<span><span>34</span>+<span><span><span><span>14</span><span>√41</span></span><span> or </span></span>x</span></span></span>=<span><span>34</span>+<span><span><span>−1</span>4</span><span>√<span>41</span></span></span></span></span>
Answer:
Given A triangle ABC in which
∠C =90°,∠A=20° and CD ⊥ AB.
In Δ ABC
⇒∠A + ∠B +∠C=180° [ Angle sum property of triangle]
⇒20° + ∠B + 90°=180°
⇒∠B+110° =180°
∠B =180° -110°
∠B = 70°
In Δ B DC
∠BDC =90°,∠B =70°,∠BC D=?
∠BDC +,∠B+∠BC D=180°[ angle sum property of triangle]
90° + 70°+∠BC D =180°
∠BC D=180°- 160°
∠BC D = 20°
In Δ AC D
∠A=20°, ∠ADC=90°,∠AC D=?
∠A + ∠ADC +∠AC D=180° [angle sum property of triangle]
20°+90°+∠AC D=180°
110° +∠AC D=180°
∠AC D=180°-110°
∠AC D=70°
So solution are, ∠AC D=70°,∠ BC D=20°,∠DB C=70°
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The answer is only I and II
APZ is the correct answer
