Updated spacing between @ latex expressions

astrosander правка от
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правка #10609 предыдущая #10120 GitHub c702ebe ← раньше
en/2.5.17.md +6 −6
@@ -75,19 +75,19 @@
Substituting into $\text{(1)}$:
$$\frac{m_2}{m_1}p_0^2-\frac{m_2}{m_1}p_1^2=p_1^2+p_0^2-2p_1p_0\cos\alpha$$
Let's regroup this into a quadratic equation
−<div class="scroll-wrapper">$$p_1^2\left(1+\frac{m_2}{m_1}\right)-2p_0p_1\cos\alpha+p_0^2\left(1-\frac{m_2}{m_1}\right)=0$$</div>
+<div class="scroll-wrapper">$$p_1^2\left(1+\frac{m_2}{m_1}\right)-2p_0p_1\cos\alpha +p_0^2\left(1-\frac{m_2}{m_1}\right)=0$$</div>
Solving relative to $p_1$
−<div class="scroll-wrapper">$$p_1=\frac{2p_0\cos\alpha\pm \sqrt{4p_0^2\cos^2 \alpha-4p_0^2\left(1-\frac{m_2^2}{m_1^2}\right)}}{2\left(1+\frac{m_2}{m_1}\right)}$$</div>
+<div class="scroll-wrapper">$$p_1=\frac{2p_0\cos\alpha\pm \sqrt{4p_0^2\cos^2 \alpha -4p_0^2\left(1-\frac{m_2^2}{m_1^2}\right)}}{2\left(1+\frac{m_2}{m_1}\right)}$$</div>
Since $\Delta = 0$
−$$\cos^2 \alpha=1-\frac{m_2^2}{m_1^2}$$
+$$\cos^2 \alpha =1-\frac{m_2^2}{m_1^2}$$
From where we find the required ratio
−$$\frac{m_2^2}{m_1^2}=1-\cos^2 \alpha=\sin^2 \alpha$$
−$$\boxed{\sin\alpha=\frac{m_2}{m_1}}$$
+$$\frac{m_2^2}{m_1^2}=1-\cos^2 \alpha =\sin^2 \alpha$$
+$$\boxed{\sin\alpha =\frac{m_2}{m_1}}$$
</p>
<h4>Answer</h4>
<p>
− $$\sin\alpha=m_2/m_1$$
+ $$\sin\alpha =m_2/m_1$$
</p>
<p style="text-align: right; font-style: italic; font-size: 14;">
ещё строк без изменений 16