A ball is thrown up from ground and its height at any time '$t$' is given by the function $h(t)$ = $-1 6t^2 + 40t$. Find the time interval in which the height of the ball is greater than $16$ feet.
As per the problem statement we need to find the time when $h(t) > 16$
In other words, $-16t^2 + 40t > 16$
Or, $-16t^2 + 40t -16 > 0$
Or, $2t^2 - 5t + 2 < 0$ (dividing throughout by $-8$ and reversing the inequality symbol)
Or, $2t(t - 2) -1(t - 2) > 0$
Or, $(2t - 1)(t - 2) > 0$
As per the negative product property, either $(2t - 1) > 0$ and $(t - 2) < 0$ or $(2t - 1) < 0$ and $(t - 2) > 0.$ But the second set of condition is Impossible because '$t$' cannot be less than $0.5$ and at the same time cannot be $> 2$. Hence, as per the first set of conditions $t > 0.5$ and $t < 2$. Therefore, the ball is at a height greater than $16$ feet between $0.5$ and $2$ seconds. The solution is graphed on a number line as shown below.
The shaded part of the number line gives the time interval of the height of the ball greater than $16$ feet.
Robert and Ben run taxis. Robert charges $ \$1.75$ fixed and $ \$0.65$ per mile whereas Ben's terms are $ \$ 2.50$ fixed and $ \$0.50$ per mile. From how many miles of journey hiring Ben's taxi will be cheaper?
Let '$x$' be the number of miles of journey and $y_1$ be the total fare in dollars with Robert's taxi and $y_2$ be that in case of Ben's taxi. Then $y_1$ = $1.75 + 0.65x$ and $y_2$ = $2.50 + 0.50x$. The given condition is, $y_2 \leq y_2$. In other words,
$2.50 + 0.50x \leq 1.75 + 0.65x$
$-0.15x \leq -0.75$
$-x \leq -5$
Or, $x \geq 5$ (note that multiplying both sides by $-1$ reverses the inequality symbol)
So, for a journey of 5 miles and above, hiring Ben's taxi will be cheaper. The solution is graphed on a number line and shown as below.
Since the answer also includes $5$ miles, there is a closed circle at $5$ on the number line.