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3 years ago

I need help with linear functions can anybody help me?

Mathematics

1 answer:

NNADVOKAT [17]3 years ago

7 0

I can help you! Just let me how what the questions are

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Simplify the expression. Write your answer as a power. <img src="https://tex.z-dn.net/?f=%285%5E%7B4%7D%20%29%5E%7B3%7D" id=

Charra [1.4K]

Answer:

$5^{12}$

Step-by-step explanation:

Lets say there's a number 'x' and there's an expression $(x^a)^b$ where 'a' & 'b' are also numbers , the simplified form of the expression is $x^{a \times b}$ .

So ,

$(5^4)^3 = 5^{4 \times 3} = 5^{12}$

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2 years ago

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3 years ago

Rectangle CDEF is rotated 45 in clockwise direction . To which angle of the image does angle D correspond? A-angle C’ B-angle D’

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3 years ago

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(a) Write 0.7 as a percentage. (b) Write 43.2% as a decimal. help me

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(a) 0.7 as a percentage is 70%

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3 years ago

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g 1) The rate of growth of a certain type of plant is described by a logistic differential equation. Botanists have estimated th

alexira [117]

Answer:

a) The expression for the height, 'H', of the plant after 't' day is;

$H = \dfrac{30}{1 + 5\cdot e^{-(2.02732554 \times 10^{-3}) \cdot t}}$

b) The height of the plant after 30 days is approximately 19.426 inches

Step-by-step explanation:

The given maximum theoretical height of the plant = 30 in.

The height of the plant at the beginning of the experiment = 5 in.

a) The logistic differential equation can be written as follows;

$\dfrac{dH}{dt} = K \cdot H \cdot \left( M - {P} \right)$

Using the solution for the logistic differential equation, we get;

$H = \dfrac{M}{1 + A\cdot e^{-(M\cdot k) \cdot t}}$

Where;

A = The condition of height at the beginning of the experiment

M = The maximum height = 30 in.

Therefore, we get;

$5 = \dfrac{30}{1 + A\cdot e^{-(30\cdot k) \cdot 0}}$

$1 + A = \dfrac{30}{5} = 6$

A = 5

When t = 20, H = 12

We get;

$12 = \dfrac{30}{1 + 5\cdot e^{-(30\cdot k) \cdot 20}}$

$1 + 5\cdot e^{-(30\cdot k) \cdot 20} = \dfrac{30}{12} = 2.5$

$5\cdot e^{-(30\cdot k) \cdot 20} = 2.5 - 1 = 1.5$

∴ -(30·k)·20 = ㏑(1.5)

k = ㏑(1.5)/(30 × 20) ≈ 6·7577518 × 10⁻⁴

k ≈ 6·7577518 × 10⁻⁴

Therefore, the expression for the height, 'H', of the plant after 't' day is given as follows

$H = \dfrac{30}{1 + 5\cdot e^{-(30\times 6.7577518 \times 10^{-4}) \cdot t}} = \dfrac{30}{1 + 5\cdot e^{-(2.02732554 \times 10^{-3}) \cdot t}}$

b) The height of the plant after 30 days is given as follows

$H = \dfrac{30}{1 + 5\cdot e^{-(2.02732554 \times 10^{-3}) \cdot t}}$

At t = 30, we have;

$H = \dfrac{30}{1 + 5\cdot e^{-(2.02732554 \times 10^{-3}) \times 30}} \approx 19.4258866473$

The height of the plant after 30 days, H ≈ 19.426 in.

3 0

3 years ago