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4 years ago

PLEASE I NEED THIS TODAY!!!

Mathematics

1 answer:

Troyanec [42]4 years ago

6 0

Answer:

Option D $y\leq -3x-3$ and $y$

Step-by-step explanation:

step 1

<u>Find the equation of the inequality A</u>

we know that

The solution of the inequality A (dashed line) is the shaded area below the dashed line

The equation of the dashed line is $y=-\frac{1}{2}x+2$

therefore

The equation of the inequality A is $y$

step 2

<u>Find the equation of the inequality B</u>

we know that

The solution of the inequality B (solid line) is the shaded area below the solid line

The equation of the solid line is $y=-3x-3$

therefore

The equation of the inequality B is $y\leq -3x-3$

The system of linear inequalities is

$y\leq -3x-3$ and $y$

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What is the answer? Please answer soon!

pshichka [43]

B because the variable (x) is not raised to any power.

4 0

3 years ago

Suppose a population grows according to the logistic equation but is subject to a constant total harvest rate of H. If N(t) is t

Elenna [48]

Answer:

a) Equilibrium point : [ 947, 53 ]

b) N = 947 is stable equilibrium, N = 53 is unstable equilibrium

c) N0, the population will not go extinct

Step-by-step explanation:

Given that;

r = 2, k = 1000, H = 100

dN/dT = R(1 - N/k)N - H

so we substitute

dN/dt = 2( 1 - N/1000)N - 100

now for equilibrium solution, dN/dt = 0

2( 1 - N/1000)N - 100 = 0

((1000 - N)/1000)N = 50

N^2 - 1000N + 50000 = 0

N = 1000 ± √(-1000)² - 4(1)(50000)) / 2(1)

N = 947.213 OR 52.786

approximately

N = 947 OR 53

Therefore Equilibrium point : [ 947, 53 ]

g(N) = 2( 1 - N/1000)N - 100

= 2N - N²/500 - 100

g'(N) = 2 - N/250

SO AT 947

g'(N) = g'(947) = 2 - 947/250 = -1.788 which is less than (<) 0

so N = 947 is stable equilibrium

now AT 53

g'(N) = g"(53) = 2 - 53/250 = 1.788 which is greater than (>) 0

so N = 53 is unstable equilibrium

The capacity k=1000

If the population is less than 53 then the population will become extinct but since the capacity is equal to 1000 then the population will not go extinct.

6 0

4 years ago

Help me plsssssssssss

prisoha [69]

Answer:

Sin 21 = p/b

0.358 = 5.5 / x

x = 5.5 / 0.358

therefore , x = 6.97

Step-by-step explanation:

Use sin formula

which is perpendicular/ base

just put the value

3 0

2 years ago

100 POINTS IF U GET THIS RIGHT!

alisha [4.7K]

Answer:

The Slope equals (-8/7)

7 0

2 years ago

Two years ago the population of a town was 40000. The population of the town at present has reached 44100. Calculate the populat

LenaWriter [7]

Answer:

5% annual population growth rate

Step-by-step explanation:

Let the percent the population grows by be $X\%$ . The total population, $f(x)$ , after $t$ years can be modeled by the function:

$f(x)=40,000\cdot (\frac{X}{100}+1)^t$

Why?

Let's take a look at a simple example. If we said a number $n$ grew by 10%, we could represent the number after it grew by multiplying $n$ by $1.10$ . This is because growing by 10% is equivalent to taking $100\%+10\%=110\%$ of that number and we convert a percentage to a decimal by dividing by 100.

Therefore, if the population grew $X\%$ , we would divide it by 100 to convert it to a decimal, then add 1 (100%) and raise to the power of $t$ (number of years) to multiply by the initial population of 40,000 to get the total population after $t$ years.

Since the population of the town after two years is 44,100, substitute $f(x)=44,100$ and $t=2$ into $f(x)=40,000\cdot (\frac{X}{100}+1)^t$ :

$44,100=40,000\cdot (\frac{X}{100}+1)^2,\\\\(\frac{X}{100}+1)^2=\frac{44,100}{40,000},\\\\(\frac{X}{100}+1)^2=1.1025,\\\\(\frac{X}{100}+1)^2=\pm \sqrt{1.1025},\\\\\begin{cases}\frac{X}{100}+1=1.05,\frac{X}{100}=0.05, X=\boxed{5\%},\\*\text{negative case is extraneous since X must be positive}\end{cases}$

Therefore, the city has an annual population growth rate of 5%.

6 0

3 years ago