When solving geometry problems with constraints, carefully analyze integer solutions and relationships between coordinates
The intercepts of the line are (20,0) and (0,15). Compute the points inside the triangle satisfying b = ka. The total is: 31points.
The correct answer is 31
\((1,1)(1,2)−(1,14)⇒14 pts.\)
If \(x=2\), \(y=\frac{27}{2} = 13.5\)
\((2,2)(2,4)…(2,12)⇒6 pts.\)
If \(x=3\), \(y=\frac{51}{4} = 12.75\)
\((3,3)(3,6)−(3,12)⇒4 pts.\)
If \(x=4\), \(y=12\)
\((4,4)(4,8)⇒2 pts.\)
If \(x=5\), \(y=\frac{45}{4} = 11.25\)
\((5,5),(5,10)⇒2 pts.\)
If \(x=6\), \(y=\frac{21}{2} = 10.5\)
\((6,6)⇒1pt\)
If \(x=7\), \(y=\frac{39}{4} = 9.75\)
(7,7)⇒1pt
If \(x=8\), \(y=9\)
\((8,8)⇒1pt\)
If \(x=9\) \(y=\frac{33}{4}= 8.25\)\(\Rightarrow\) no pt.
Total = 31 pts.
In the given figure, the blocks $A$, $B$ and $C$ weigh $4\,\text{kg}$, $6\,\text{kg}$ and $8\,\text{kg}$ respectively. The coefficient of sliding friction between any two surfaces is $0.5$. The force $\vec{F}$ required to slide the block $C$ with constant speed is ___ N.
(Given: $g = 10\,\text{m s}^{-2}$) 
Two circular discs of radius \(10\) cm each are joined at their centres by a rod, as shown in the figure. The length of the rod is \(30\) cm and its mass is \(600\) g. The mass of each disc is also \(600\) g. If the applied torque between the two discs is \(43\times10^{-7}\) dyne·cm, then the angular acceleration of the system about the given axis \(AB\) is ________ rad s\(^{-2}\).

A polynomial that has two roots or is of degree 2 is called a quadratic equation. The general form of a quadratic equation is y=ax²+bx+c. Here a≠0, b, and c are the real numbers.
Consider the following equation ax²+bx+c=0, where a≠0 and a, b, and c are real coefficients.
The solution of a quadratic equation can be found using the formula, x=((-b±√(b²-4ac))/2a)
Read More: Nature of Roots of Quadratic Equation