All categories

3 years ago

What is 2 and 2/3 mi in feet

Mathematics

2 answers:

Nina [5.8K]3 years ago

7 0

14080 ft is the answer

hope this helps you

Nuetrik [128]3 years ago

6 0

The answer is 14080 ft

You might be interested in

Please help... I really need this.

valentinak56 [21]

Answer:

$\frac{4x^3-3x^2+5x+6}{x+6}=4x^2-27x+167-\frac{996}{x+6}$

Step-by-step explanation:

The division problem given to us is $(4x^3-3x^2+5x+6)\div (x+6)$ .

To perform the synthetic division, we write out the coefficient of the polynomial. We set the linear factor to zero and solve for x, this becomes our divisor. That is $x+6=0\implies x=-6$ .

We carry out the synthetic division as shown in the attachment.

The result of the synthetic division is 4 -27 167 -996

The first three terms are the coefficients of the quotient and the last term is the remainder.

Therefore the quotient is $q(x)=4x^2-27x+167$ and the remainder is $r(x)=-996$ .

The dividend, the quotient and the divisor are written as

$\frac{4x^3-3x^2+5x+6}{x+6}=4x^2-27x+167-\frac{996}{x+6}$

The correct answer is A

3 0

3 years ago

Question 4

g100num [7]

A, 3x5=15 and 3x8=24

4 0

3 years ago

Read 2 more answers

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

The interquartile range of this data set

Natali [406]

The interquartile range is 20 you take Q3 in the graph which is 30 and the Q1 in the graph which is 10 then you subtract Q3-Q1 so 30-10 and get 20

6 0

2 years ago

Read 2 more answers

Help me please if i dont do my homework my mom will beat me

Snezhnost [94]

3.5 is the answer

I did the math and that is what i got.

4 0

3 years ago

Read 2 more answers