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A big ship drops its anchor.

E represents the anchor's elevation relative to the water's surface (in meters) as a function of time
t (in seconds).

E=-2.4 t+75
How far does the anchor drop every 5 seconds?
meters

A big ship drops its anchor. $\newline$ $E$ represents the anchor's elevation relative to the water's surface (in meters) as a function of time $t$ (in seconds). $\newline$ $E=-2.4 t+75$ $\newline$ How far does the anchor drop every $5$ seconds? $\newline$ meters

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Write exponential functions: word problems

Full solution

Q. A big ship drops its anchor.

\newline

E

represents the anchor's elevation relative to the water's surface (in meters) as a function of time

t

(in seconds).

\newline

E=-2.4 t+75

\newline

How far does the anchor drop every

5

seconds?

\newline

meters

Calculate initial elevation: To find out how far the anchor drops every ext{ extdollar} $5$ ext{ extdollar} seconds, we need to calculate the change in elevation after ext{ extdollar} $5$ ext{ extdollar} seconds using the given equation ext{ extdollar}E = $-2$ . $4$ t + $75$ ext{ extdollar}.
Calculate elevation at extdollar{} $5$ extdollar{} seconds: First, we calculate the elevation at extdollar{}t = $0$ extdollar{} seconds, which is the initial elevation before the anchor starts to drop. $\newline$ extdollar{}E( $0$ ) = $-2$ . $4$ ( $0$ ) + $75$ = $75$ extdollar{} meters.
Find difference in elevation: Next, we calculate the elevation at $t = 5$ seconds. $\newline$ $E(5) = -2.4(5) + 75 = -12 + 75 = 63$ meters.
Determine drop distance: Now, we find the difference in elevation between $t = 0$ seconds and $t = 5$ seconds to determine how far the anchor has dropped. $\newline$ Drop distance = $E(0) - E(5) = 75 \text{ meters} - 63 \text{ meters} = 12 \text{ meters}.$

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