(20 POINTS AND BRAINLIEST ANSWER!!!) I need help with these two Geometry 1 questions!

Write the equation of a line that is parallel to y = 1/2 x + 5

kumpel [21]

Answer:

y = 1/2x + 5

Step-by-step explanation:

Slope-Intercept Form: y = mx + b

When 2 lines are parallel, they have the same slope.

Step 1: Define variables

Random point (2, 6)

<em>m</em> = 1/2

y = 1/2x + b

Step 2: Find <em>b</em>

6 = 1/2(2) + b

6 = 1 + b

5 = b

Step 3: Rewrite parallel linear equation

y = 1/2x + 5

3 0

3 years ago

A forestry company hopes to generate credits from the carbon sequestered in its plantations.The equation describing the forest w

Dmitry [639]

Answer:

Explanation:

Given:

The equation describing the forest wood biomass per hectare as a function of plantation age t is:

y(t) = 5 + 0.005t^2 + 0.024t^3 − 0.0045t^4

The equation that describes the annual growth in wood biomass is:

y ′ (t) = 0.01t + 0.072t^2 - 0.018t^3

To find:

a) The year the annual growth achieved its highest possible value

b) when does y ′ (t) achieve its highest value?

a)

To determine the year the highest possible value was achieved, we will set the derivative y'(t) to zero. The values of t will be substituted into the second derivative to get the highest value

$\begin{gathered} 0\text{ = 0.01t + 0.072t}^2\text{ - 0.018t}^3 \\ using\text{ a graphing tool,} \\ t\text{ = -0.1343} \\ \text{t = 0} \\ t\text{ = 4.1343} \\ \\ Since\text{ we can't have time as a negative value, t = 0 or t = 4.1343} \end{gathered}$ $\begin{gathered} To\text{ ascertain which is the maximum, we take a second derivative} \\ y^{\prime}\text{'\lparen t\rparen = 0.01 + 0.144t - 0.054t}^2 \\ \\ substitute\text{ t = 0 and t = 4.13 respectively} \\ when\text{ t = 0 } \\ y^{\prime}\text{'\lparen t\rparen= 0.01 + 0.144\lparen0\rparen - 0.054\lparen0\rparen}^2 \\ y^{\prime}\text{'\lparen t\rparen= 0.01 + 0 - 0} \\ y^{\prime}\text{'\lparen t\rparen= 0.01 > 0} \\ \\ y^{\prime}\text{'\lparen t\rparen= 0.01 + 0.144\lparen4.13\rparen - 0.054\lparen4.13\rparen}^2 \\ y^{\prime}\text{'\lparen t\rparen= -0.316 < 0} \\ \\ if\text{ }y^{\prime}\text{'\lparen t\rparen > 0 , then it is minimum, } \\ if\text{ }y^{\prime}\text{'\lparen t\rparen < 0, then it is maximum} \end{gathered}$

SInce t = 4.13, gives y ′' (t) = -0.316 (< 0). This makes it the maximum value of t

The year the annual growth achieved its highest possible value to the nearest whole number will be

year 4

b) y ′ (t) will achieve its highest value, when we substitute the value of t that gives into the initial function.

Initial function: y(t) = 5 + 0.005t^2 + 0.024t^3 − 0.0045t^4