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miss Akunina [59]

2 years ago

9

A newly drilled water well produces 95,000 quarts of water per week. With no new water feeding the well, the production drops by

per year. Using 52 weeks in a year, what is the total number of quarts of water that can be drawn from this water well before it goes dry?

1 answer:

VARVARA [1.3K]2 years ago

5 0

Step-by-step explanation:

1 week =95000

52 weeks =x

x=95000×52

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For the following system, determine if a steady state exists and give the steady state value. The population of a colony of squi

Anettt [7]

Answer:

Option A)

$\displaystyle\lim_{t\to\infty}~p(t) = k$

Step-by-step explanation:

We are given the following on the question:

$p(t) =\displaystyle\frac{4000}{4+e^{-0.02t}}$

where p(t) is the population of a colony.

For steady state solution we evaluate:

$\displaystyle\lim_{t\to\infty}~p(t)\\\\= \lim_{x\to\infty} \frac{4000}{4+e^{-0.02t}}\\\\=\frac{4000}{4+e^{-\infty}}\\\\=\frac{4000}{4}\\\\= 1000$

Thus, the steady state solution is a constant, k = 1000.

Thus, the correct answer is

Option A)

$\displaystyle\lim_{t\to\infty}~p(t) = k$

8 0

2 years ago

Need Help which expression is equal to

a_sh-v [17]

Answer:

4^0

Step-by-step explanation:

whenever the number is to the power of 0 it equals 1

4^5 x 4^-7 ÷ 4^-2 = 1

4^0 = 1

3 0

2 years ago

Use the given information to find the exact value of the trigonometric function

eimsori [14]

$\begin{gathered} \csc \theta=-\frac{6}{5} \\ \tan \theta>0 \\ \cos \frac{\theta}{2}=\text{?} \end{gathered}$

Half Angle Formula

$\cos \frac{\theta}{2}=\pm\sqrt[\square]{\frac{1+\cos\theta}{2}}$ $\tan \theta>0\text{ and csc}\theta\text{ is negative in the third quadrant}$ $\begin{gathered} \csc \theta=-\frac{6}{5}=\frac{r}{y} \\ x^2+y^2=r^2 \\ x=\pm\sqrt[\square]{r^2-y^2} \\ x=\pm\sqrt[\square]{6^2-(-5)^2} \\ x=\pm\sqrt[\square]{36-25} \\ x=\pm\sqrt[\square]{11} \\ \text{x is negative since the angle is on the 3rd quadrant} \end{gathered}$ $\begin{gathered} \cos \theta=\frac{x}{r}=\frac{-\sqrt[\square]{11}}{6} \\ \cos \frac{\theta}{2}=\pm\sqrt[\square]{\frac{1+\cos\theta}{2}} \\ \cos \frac{\theta}{2}is\text{ also negative in the 3rd quadrant} \\ \cos \frac{\theta}{2}=-\sqrt[\square]{\frac{1+\frac{-\sqrt[\square]{11}}{6}}{2}} \\ \cos \frac{\theta}{2}=-\sqrt[\square]{\frac{\frac{6-\sqrt[\square]{11}}{6}}{2}} \\ \cos \frac{\theta}{2}=-\sqrt[\square]{\frac{6-\sqrt[\square]{11}}{12}} \\ \\ \end{gathered}$

Answer:

$\cos \frac{\theta}{2}=-\sqrt[\square]{\frac{6-\sqrt[\square]{11}}{12}}$

Checking:

$\begin{gathered} \frac{\theta}{2}=\cos ^{-1}(-\sqrt[\square]{\frac{6-\sqrt[\square]{11}}{12}}) \\ \frac{\theta}{2}=118.22^{\circ} \\ \theta=236.44^{\circ}\text{ (3rd quadrant)} \end{gathered}$

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The answer is none of the choices

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4 months ago

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Lilit [14]

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3 years ago

In angle PQR, angle P = (4x – 14)º, angle Q = (5x + 6)", and angle R= (x - 2)°. Find<br> angle R.

vekshin1

we have: ∠P + ∠Q + ∠R = 180°

<=> (4x – 14)º + (5x + 6)° + (x - 2)° = 180°

<=> 4x – 14 + 5x + 6 + x - 2 = 180°

<=> 10x - 10 = 180°

<=> 10x = 190

<=> x = 19°

So: ∠R = (x - 2)° = (19 - 2)° = 17°

ok done. Thank to me :>

5 0

1 year ago