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4 years ago

12

Write a number that has four digits with the same number in all places,such as 4,444. Circle the digits with the greatest value.

Underline the digit with the least value.Explain

1 answer:

Citrus2011 [14]4 years ago

6 0

9999

first 3 nines have the greater value but the first one has the highest and the least one is the last nine.

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To prove :One plus cot square theta into tan theta by sec square theta = cot theta

pickupchik [31]

we are given

$\frac{(1+cot^2(\theta))*tan(\theta)}{sec^2(\theta)} =cot(\theta)$

We will simplify left side and make it equal to right side

Left side:

$\frac{(1+cot^2(\theta))*tan(\theta)}{sec^2(\theta)}$

we can use trigonometric identity

$1+cot^2(\theta)=csc^2(\theta)$

we can replace it

$\frac{(csc^2(\theta))*tan(\theta)}{sec^2(\theta)}$

we know that

csc=1/sin and sec=1/cos

so, we can replace it

and we get

$\frac{cos^2(\theta)tan(\theta)}{sin^2(\theta)}$

now, we know that

tan =sin/cos

$\frac{cos^2(\theta)*sin(\theta)}{sin^2(\theta)*cos(\theta)}$

we can simplify it

and we get

$\frac{cos(\theta)}{sin(\theta)}$

we can also write it as

$=cot(\theta)$

Right Side:

$cot(\theta)$

we can see that

left side = right side

so,

$\frac{(1+cot^2(\theta))*tan(\theta)}{sec^2(\theta)} =cot(\theta)$ ......Answer

6 0

3 years ago

What two numbers multiply to get -2 and then add to get 3?

lana [24]

If you got options that will help 1×-2=-2 1+2=3

6 0

3 years ago

A corporation with several thousand employees wants to estimate the mean commute time for all employees. They would like to cons

velikii [3]

Answer:

C.25

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.95}{2} = 0.025$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of $1-0.025 = 0.975$ , so $z = 1.96$

Now, find the margin of error M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

Which of the following is the smallest sample the company can take to achieve the desired margin or error?

This is n when $\sigma = 10, M = 4$

$M = z*\frac{\sigma}{\sqrt{n}}$

$4 = 1.96*\frac{10}{\sqrt{n}}$

$4\sqrt{n} = 1.96*10$

$\sqrt{n} = \frac{1.96*10}{4}$

$(\sqrt{n})^{2} = (\frac{1.96*10}{4})^{2}$

$n = 24.01$

We have to round up.

So the correct answer is:

C.25

3 0

3 years ago

The logistic equation for the population (in thousands) of a certain species is given by:

Eva8 [605]

Answer:

a.

b. 1.5

c. 1.5

d. No

Step-by-step explanation:

a. First, let's solve the differential equation:

$\frac{dp}{dt} =3p-2p^2$

Divide both sides by $3p-2p^2$ and multiply both sides by dt:

$\frac{dp}{3p-2p^2}=dt$

Integrate both sides:

$\int\ \frac{1}{3p-2p^2} dp =\int\ dt$

Evaluate the integrals and simplify:

$p(t)=\frac{3e^{3t} }{C_1+2e^{3t}}$

Where C1 is an arbitrary constant

I sketched the direction field using a computer software. You can see it in the picture that I attached you.

b. First let's find the constant C1 for the initial condition given:

$p(0)=3=\frac{3e^{0} }{C_1+2e^{0} } =\frac{3}{C_1+2}$

Solving for C1:

$C_1=-1$

Now, let's evaluate the limit:

$\lim_{t \to \infty} \frac{3e^{3t} }{2e^{3t}-1 } \\\\Divide\hspace{3}the\hspace{3}numerator\hspace{3}and\hspace{3}denominator\hspace{3}by\hspace{3}e^{3t} \\\\ \lim_{t \to \infty} \frac{3 }{2-e^{-3x} }$

The expression $-e^{-3x}$ tends to zero as x approaches ∞ . Hence:

$\lim_{t \to \infty} \frac{3e^{3t} }{2e^{3t}-1 } =\frac{3}{2} =1.5$

c. As we did before, let's find the constant C1 for the initial condition given:

$p(0)=0.8=\frac{3e^{0} }{C_1+2e^{0} } =\frac{3}{C_1+2}$

Solving for C1:

$C_1=1.75$

Now, let's evaluate the limit:

$\lim_{t \to \infty} \frac{3e^{3t} }{2e^{3t}+1.75 } \\\\Divide\hspace{3}the\hspace{3}numerator\hspace{3}and\hspace{3}denominator\hspace{3}by\hspace{3}e^{3t} \\\\ \lim_{t \to \infty} \frac{3 }{2+1.75e^{-3x} }$

The expression $-e^{-3x}$ tends to zero as x approaches ∞ . Hence:

$\lim_{t \to \infty} \frac{3e^{3t} }{2e^{3t}+1.75 } =\frac{3}{2} =1.5$

d. To figure out that, we need to do the same procedure as we did before. So, let's find the constant C1 for the initial condition given:

$p(0)=2=\frac{3e^{0} }{C_1+2e^{0} } =\frac{3}{C_1+2}$

Solving for C1:

$C_1=-\frac{1}{2} =-0.5$

Can a population of 2000 ever decline to 800? well, let's find the limit of the function when it approaches to ∞:

$\lim_{t \to \infty} \frac{3e^{3t} }{2e^{3t}-0.5 } \\\\Divide\hspace{3}the\hspace{3}numerator\hspace{3}and\hspace{3}denominator\hspace{3}by\hspace{3}e^{3t} \\\\ \lim_{t \to \infty} \frac{3 }{2-0.5e^{-3x} }$

The expression $-e^{-3x}$ tends to zero as x approaches ∞ . Hence:

$\lim_{t \to \infty} \frac{3e^{3t} }{2e^{3t}-0.5 } =\frac{3}{2} =1.5$

Therefore, a population of 2000 never will decline to 800.

6 0

3 years ago

• 2 * 3 = 15<br><br>3 * 4 = 28<br><br>4 * 5 = 45<br><br>Then, 6 * 2 = _______

olga_2 [115]

Answer:

16?

Step-by-step explanation:

6 0

3 years ago