A body of mass 2kg is projected from the ground with a velocity $20m{s}^{-1}$ at an angle ${30}^{\circ }$ with the verticle. If $t_1$ is the time in seconds at which the body is projected and $t_2$ is the time in seconds at which it reaches the ground, the change in momentum in $kgm{s}^{-1}$ during the time $\left(t_2-t_1\right)$ is a. 40 b. $40\sqrt{3}$ c. $50\sqrt{3}$ d. 60

Sure, here are some step-by-step solutions: ## Example 1: Simplify the expression: $3x^2-2x^2+5x-4x$ **Solution:** $3x^2-2x^2$ simplifies to $x^2$, and $5x-4x$ simplifies to $x$. Therefore, the expression simplifies to: $\boxed{x^2+x}$ ## Example 2: Solve the equation: $2x+5=13$ **Solution:** To isolate the variable $x$, we can subtract 5 from both sides of the equation: $\begin{array}{rl}2x+5-5& =13-5\\ 2x& =8\end{array}$ Then, divide both sides of the equation by 2: $\begin{array}{rl}\frac{2x}{2}& =\frac{8}{2}\\ x& =4\end{array}$ Therefore, the solution is: $\boxed{x=4}$ ## Example 3: Factor the expression: $6x^2-11x-10$ **Solution:** To factor this quadratic expression, we need to find two numbers that multiply to give $6\cdot (-10)=-60$, and add to give $-11$. After some trial and error, we can see that $-15$ and $4$ satisfy these conditions. Therefore, we can write the expression as the product of two binomials: $\boxed{(2x-5)(3x+2)}$ ## Example 4: Solve the system of equations: $\begin{array}{rl}y& =2x-1\\ y& =x+3\end{array}$ **Solution:** Since both equations are equal to $y$, we can set them equal to each other: $\begin{array}{rl}2x-1& =x+3\\ x& =4\end{array}$ Then, we can substitute this value of $x$ into one of the original equations to find $y$: $\begin{array}{rl}y& =2(4)-1\\ y& =7\end{array}$ Therefore, the solution is: $\boxed{(x,y)=(4,7)}$ These step-by-step solutions should help students understand how to approach and solve various problems.