Question

eoc style question
kay was walking along the bridge above the river. she accidentally lost her grip on her camera when she was trying to get it out to photograph massive bird nests in the trees. the height of the camera above the water can be modeled by the equation $y = -16t^2 + 400$, where $t$ represents the seconds after the camera is dropped. how long will it take before kay sees the splash from the camera hitting the water?
\boxed{} seconds
eoc style question
what is the solution(s) to the equation $5a^2 - 44 = 81$?
\circ $\pm25$
\circ $-5$
\circ $\pm5$
\circ $\pm125$

Explanation:

First Question (Camera Dropping Problem)

Step1: Determine when the camera hits the water.

When the camera hits the water, the height \( y = 0 \). So we set the equation \( -16t^{2}+400 = 0 \).

Step2: Solve for \( t \).

First, subtract 400 from both sides: \( -16t^{2}=- 400 \).
Then, divide both sides by - 16: \( t^{2}=\frac{-400}{-16}=25 \).
Take the square root of both sides: \( t=\pm\sqrt{25}=\pm5 \). Since time cannot be negative in this context, we take \( t = 5 \).

Step1: Isolate the squared term.

Start with the equation \( 5a^{2}-44 = 81 \). Add 44 to both sides: \( 5a^{2}=81 + 44=125 \).

Step2: Solve for \( a^{2} \) and then \( a \).

Divide both sides by 5: \( a^{2}=\frac{125}{5} = 25 \).
Take the square root of both sides: \( a=\pm\sqrt{25}=\pm5 \).

Answer:

5

Second Question (Equation Solution)