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ss7ja [257]
3 years ago
13

The question is below.

Mathematics
1 answer:
svetlana [45]3 years ago
3 0

Answer:

Product before increasing: 4200

x after increase: 60 * 105% = 63

y after increase: 70 * 110% = 77

Product after increasing: 4851

Percentage increase in the product: 4851 / 4200 * 100 = 115.5%

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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?

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

undefined

6 0
2 years ago
Ramona deposited $4,190.51 into a savings account with an interest rate of 5.2% compounded twice a year. About how long will it
kap26 [50]
<span>A = P(1+r/2)^2t A/P = (1+r/2)^2t ln(A/P) = 2t ln((1+r/2)) ln(A/P)/ln(1+r/2) =2t ln(9000/4190.51)/ln(1+0.052/2)=2t t=15 yrs therfore ans is C.14 years ,11 months</span>
7 0
3 years ago
Solve by factoring X^2-100=0
Greeley [361]
X= -10, 10


This is the answer.
3 0
3 years ago
What is the value of x?
Helga [31]

2x+20=3x-30

x=50

2x+20=3x-30

x=50

2x+20=3x-30

x=50

2x+20=3x-30

x=50

7 0
3 years ago
If z varies directly as x and z=30 when x=8 find z when x=4
dangina [55]
Z=kx, 30=8k, so k=30/8=15/4 and z=15x/4, so when x=4, z=15.
8 0
3 years ago
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