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1 year ago

Simplify the following expression (m3)÷m18

Mathematics

1 answer:

faust18 [17]1 year ago

5 0

Answer:

$\boxed{ \bold{ \frac{1}{ {a}^{15} } }}$

Step-by-step explanation:

$({m}^{3} ) \: \div \: {m}^{18}$

We remove unnecessary parentheses.

${m}^{3} \: \div \: {m}^{18}$

We calculate the quotient (it is negative).

${m}^{ - 15}$

We transform the term into positive with this expression:

$\boxed{ \boxed{{a}^{ - n} \: = \: \frac{1}{ {a}^{n} } }}$

______________________

<h3>We apply the expression: </h3>

$\boxed{ \bold{ \frac{1}{ {a}^{15} } }}$

<h3><em><u>MissSpanish</u></em></h3>

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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?

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

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6 0

1 year ago

Please answer this correct answer now fast

lutik1710 [3]

Answer:

WX = 8 mm

Step-by-step explanation:

To be able to solve for WX, we need to first find the size of angle $\angle z$ .

We use the law of sines in the blue triangle to do such:

$\frac{sin(z)}{11} =\frac{sin(133)}{20} \\sin(z)=\frac{11\,sin(133)}{20} \\sin(z)=0.4022$

Now we can use this value in the larger right angle triangle where WX is the opposite side to angle $\angle z$ , and the 20 mm side is the hypotenuse:

$sin(z)=\frac{opposite}{hypotenuse} \\sin(z)=\frac{WX}{20}\\0.4022=\frac{WX}{20}\\WX=20\,(0.4022)\\WX=8.044\,\,mm$