Noah is studying a type of sea sponge that grows at a constant rate. They measured the age and height of pieces of the sponge. Here's what they found: A graph plots a=age in years, from 0 to 12, in increments of 2, versus h=height in centimeters, from 0 to 24, in increments of 4, on a coordinate plane. Points are plotted as follows: (4, 8), (8, 16), and (10, 20). Noah wants an equation they can use to find the age, in years, ( ) of any piece of this sponge based on its height in centimeters ( ). Complete Noah's equation.



Answer :

Answer:

Step-by-step explanation:

To find an equation that relates the age \( a \) of the sea sponge (in years) to its height \( h \) (in centimeters), we can use the points provided: \( (4, 8), (8, 16), (10, 20) \).

These points suggest a linear relationship between age and height. We can find the slope \( m \) of the line using two points, for example, \( (4, 8) \) and \( (10, 20) \):

\[ m = \frac{20 - 8}{10 - 4} = \frac{12}{6} = 2 \]

Now that we have the slope \( m = 2 \), we can use the point-slope form of the equation of a line:

\[ h - h_1 = m(a - a_1) \]

Let's use point \( (4, 8) \) to find the equation:

\[ h - 8 = 2(a - 4) \]

Now, solve for \( a \):

\[ h - 8 = 2a - 8 \]

\[ h = 2a \]

Therefore, the equation that relates the age \( a \) of the sea sponge to its height \( h \) is \( \boxed{a = \frac{h}{2}} \).

This equation indicates that the age \( a \) of the sponge is half of its height \( h \) in centimeters.

japtc

Answer:

The equation that Noah can use to find the age (a) of the sea sponge based on its height (h) in centimeters is:

a = 0.5h

This equation can be derived from the given information:

- The three data points provided are (4, 8), (8, 16), and (10, 20).

- These points show that as the age (a) increases by 2 years, the height (h) increases by 4 cm.

- Therefore, the rate of growth is 2 cm per year, or 0.5 years per cm.

- This linear relationship can be expressed as the equation a = 0.5h, where a is the age in years and h is the height in centimeters.