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2 years ago

i need to find an equation for this problem that applies to any step (e.g. step 100) but my brain doesn't work

Mathematics

2 answers:

Anton [14]2 years ago

6 0

Answer:

its

Step-by-step explanation:

Tomtit [17]2 years ago

5 0

I think the answer would be Y=3x+1

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Which statement is true about the expression?

zysi [14]

Answer:

Your number is (3 sqrt(2)) / sqrt(2) = 3, and is a rational number indeed. I don't know exactly how to interpret the rest of the question. If r is a positive rational number and p is some positive real number, then sqrt(r^2 p) / sqrt(p) is always rational, being equal r. Possibly your question refers to situtions in which sqrt(c) is not uniquely determined, as for c negative real number or complex non-real number. In those situations a discussion is necessary. Also, in general expressions the discussion is necesary, because the denominator must be different from 0, and so on.

Step-by-step explanation:

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2 years ago

What is the value of two numbers whose sum is 17 and their difference is 3

marysya [2.9K]

X will equal 10 and y will equal 7

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3 years ago

Lisa and experience shipping clerk can file a certain order in 2 hours. Jim a new clerk needs 3 hours to do the same job working

Neporo4naja [7]

Answer:

Together they will complete the order in 6/5 hours.

Step-by-step explanation:

Lisa an experience shipping clerk can fill a certain order in 2 hours.

Jim a new clerk needs 3 hours to do the same job working.

For 1 hours Lisa can fill 1/2 a order.

For 1 hours Jim can fill 1/3 a order.

For 1 hours togethers can fill 1/2+ 1/3=5/6 a order.

We have the following proportion:

$1:\frac{5}{6}=x:1\\\\\frac{5}{6}x=1\\\\x=\frac{6}{5}$

Together they will complete the order in 6/5 hours.

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3 years ago

Based on the data below, predict the number of square kilometers of land that will be occupied by inch worms in 50 years.

Feliz [49]

Answer:

After 50 years, the inch worms will occupy about 1,709,102 square kilometers.

6 0

2 years ago

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

Eva8 [605]

Answer:

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

2 years ago