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3 years ago

Is 74.3 grater than or less than 7.43

Mathematics

2 answers:

melamori03 [73]3 years ago

8 0

74.3 is greater than 7.43

iris [78.8K]3 years ago

3 0

It’s greater since it’s 74.3 compared to 7.43
74.3 is a larger number then 7.43

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Flauer [41]

Answer:

vertex is a point where two sides meet

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1 year 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?

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 help, basic Algebra question

Tju [1.3M]

Answer:

The given expression $(\frac{5}{6} a^{9}p^{5}) ^{3} = \frac{125}{216} \times a^{(27) } \times p^{(15)}$

Step-by-step explanation:

Here, the given expression is: $(\frac{5}{6} a^{9}p^{5}) ^{3}$

Now, starting from the outer most bracket.

As we know :

$(abc)^{n} = (a)^{n} \times (b)^{n} \times (c)^{n}$

and $(a^m)^{n} = a^{(m \times n)}$

⇒ $(\frac{5}{6} a^{9}p^{5}) ^{3} = (\frac{5}{6})^{3} \times (a^{9})^{3} \times (p^{5}) ^{3}\\$

$=\frac{125}{216} \times (a)^{(9\times3) } \times (p)^{(5 \times 3)}\\= \frac{125}{216} \times a^{(27) } \times p^{(15)}$

Hence, the given expression $(\frac{5}{6} a^{9}p^{5}) ^{3} = \frac{125}{216} \times a^{(27) } \times p^{(15)}$

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2 years ago

Read 2 more answers

What is the cost of 2.6 pounds at $6.75 per pound?

lisabon 2012 [21]

Since the cost is $6.75 per pound, and we have 2.6 pounds, multiply 6.75•2.6 to find the cost.
It would cost $17.55.

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3 years ago

Write an expression based on description; use grouping symbols

Harman [31]

Answer:

it's your answer and make me barinly least

8 0

2 years ago