Question

the height of a diver above the water, is given by $h(t) = -5t^2 + 10t + 3$, where $t$ is time measured in seconds and $h(t)$ is measured in meters. select all statements that are true about the situation.\
a:\
the diver begins 5 meters above the water.\
\
b:\
the diver begins 3 meters above the water.\
\
c:\
the function has 1 zero that makes sense in this situation.\
\
d:\
the function has 2 zeros that make sense in this situation.\
\
e:\
the graph that represents $h$ starts at the origin and curves upward.\
\
f:\
the diver begins at the same height as the water level.

Explanation:

Step1: Analyze the initial height (t=0)

To find the initial height, substitute \( t = 0 \) into \( h(t)=-5t^{2}+10t + 3 \).
\( h(0)=-5(0)^{2}+10(0)+3=3 \). So the diver begins 3 meters above the water. Thus, statement B is true, A and F are false.

Step2: Analyze the zeros of the function

We need to solve \( -5t^{2}+10t + 3 = 0 \). Multiply both sides by -1 to get \( 5t^{2}-10t - 3=0 \).
Using the quadratic formula \( t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a} \), where \( a = 5 \), \( b=- 10 \), \( c=-3 \).
First, calculate the discriminant \( \Delta=b^{2}-4ac=(-10)^{2}-4\times5\times(-3)=100 + 60 = 160 \).
Then \( t=\frac{10\pm\sqrt{160}}{10}=\frac{10\pm4\sqrt{10}}{10}=\frac{5\pm2\sqrt{10}}{5} \).
We have two solutions: \( t_1=\frac{5 + 2\sqrt{10}}{5}\approx\frac{5+6.32}{5}\approx2.26 \) and \( t_2=\frac{5-2\sqrt{10}}{5}\approx\frac{5 - 6.32}{5}\approx - 0.26 \).
Since time \( t\geq0 \) in this context, only \( t_1 \) is valid. So the function has 1 zero that makes sense. Thus, statement C is true, D is false.

Step3: Analyze the graph of the function

The function \( h(t)=-5t^{2}+10t + 3 \) is a quadratic function with \( a=-5<0 \), so the parabola opens downward. When \( t = 0 \), \( h(0)=3
eq0 \), so the graph does not start at the origin. Thus, statement E is false.

Answer:

B. The diver begins 3 meters above the water.
C. The function has 1 zero that makes sense in this situation.