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3 years ago

Please help don't understand

Mathematics

1 answer:

rewona [7]3 years ago

5 0

-1.5(4-n)+2.8

multiply 4-n by -1.5
multiply to both the 4 and -n

-6+1.5n+2.8

add -6+2.8 since they are similar

-3.2+1.5n

1.5n-3.2

I believe this is the simplest form

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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?

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

$undefined$

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1 year ago

Math help Please!!!!

GalinKa [24]

Part A. What is the slope of a line that is perpendicular to a line whose equation is −2y=3x+7?

Rewrite the equation −2y=3x+7 in the form $y=-\dfrac{3}{2}x-\dfrac{7}{2}.$ Here the slope of the given line is $m_1=-\dfrac{3}{2}.$ If $m_2$ is the slope of perpendicular line, then

$m_1\cdot m_2=-1,\\ \\m_2=-\dfrac{1}{m_1}=\dfrac{2}{3}.$

Answer 1: $\dfrac{2}{3}$

Part B. The slope of the line y=−2x+3 is -2. Since $-\dfrac{3}{2}\neq -2\quad \text{and}\quad \dfrac{2}{3}\neq -2,$ then lines from part A are not parallel to line a.

Since $-2\cdot \left(-\dfrac{3}{2}\right)=3\neq -1\quad \text{and}\quad -2\cdot \dfrac{2}{3}=-\dfrac{4}{3}\neq -1,$ both lines are not perpendicular to line a.

Answer 2: Neither parallel nor perpendicular to line a

Part C. The line parallel to the line 2x+5y=10 has the equation 2x+5y=b. This line passes through the point (5,-4), then

2·5+5·(-4)=b,

10-20=b,

b=-10.

Answer 3: 2x+5y=-10.

Part D. The slope of the line $y=\dfrac{x}{4}+5$ is $\dfrac{1}{4}.$ Then the slope of perpendicular line is -4 and the equation of the perpendicular line is y=-4x+b. This line passes through the point (2,7), then