2.8x-1.6y=3.2<br> -0.7x+1.9y=6.7

A species of horned beetle has a (starting) population size of 4000 individuals. Every month, 170 individuals are born. However,

erma4kov [3.2K]

Answer:

Kindly check explanation

Step-by-step explanation:

Given the following :

Starting population = 4000

Addition per month = 170

decline on population per month = 70

Increase rate in population per month (dt) :

Starting population = 4000

Number of births per month = 170

However, the population declines by 70 individuals each month

Hence,

Number of births - number of deaths(d) = 70

170 - d = - 70 ( decline?

170 + 70 = d

240 = d

d = number of deaths

Per capita death :

Total number of deaths per. Month / starting population

= 240 / 4000

= 0.06

8 0

1 year ago

Find the limits of integration ly, uy, lx, ux, lz, uz (some of which will involve variables x,y,z) so that ∫uyly∫uzlz∫uxlxdxdzdy

motikmotik

Answer:

Hello your question is incomplete attached below is the complete question

Ix = 0 Ux = $\sqrt{y-9z^2}$

Iz = 0 Uz = $\frac{\sqrt{y} }{3}$

Iy = 5 Uy = 10

Step-by-step explanation:

Ix = 0 Ux = $\sqrt{y-9z^2}$

Iz = 0 Uz = $\frac{\sqrt{y} }{3}$

Iy = 5 Uy = 10

attached below is the detailed solution

4 0

1 year ago

Evaluate the step function for the given input values. g(x) = StartLayout enlarged left-brace 1st row 1st column negative 4, 2nd

Ostrovityanka [42]

Answer:

g(2) = 3

g(–2) = -4

g(5) = 5

Step-by-step explanation:

5 0

1 year ago

Read 2 more answers

In a café, drinks are priced according to what size they are.

Andrei [34K]

A) Gavin’s drinks altogether cost £9.00
1.50+2.50x3
=1.50+7.50
=£9.00
b) Lian gets back £3.50
1.50+2.00+3.00
=6.50

10.00-6.50=£3.50

8 0

2 years ago

Himpunan penyelesaian pertidaksamaan nilai mutlak |2x-4| > |3x+2|

Stells [14]

Answer:

The solution in interval notation is:

$(-6,\frac{2}{5})$ .

The solution in inequality notation is:

$-6$ .

Step-by-step explanation:

I think you are asking how to solve this for $x$ .

Keep in mind $|x|=\sqrt{x^2}$ .

$|2x-4|>|3x+2|$

$\sqrt{(2x-4)^2}>\sqrt{(3x+2)^2}$

If $\sqrt{u}>\sqrt{v}$ then $u>v$ .

$(2x-4)^2>(3x+2)^2$

Subtract $(2x-4)^2$ on both sides:

$0>(3x+2)^2-(2x-4)^2$

Factor the difference of squares $a^2-b^2=(a-b)(a+b)$ :

$0>((3x+2)-(2x-4))((3x+2)+(2x-4))$

Simplify inside the factors:

$0>(x+6)(5x-2)$

$(x+6)(5x-2)$

The left hand side is a parabola that faces up. I know this because the degree is 2.

The zeros of the the parabola are at x=-6 and x=2/5.

We can solve x+6=0 and 5x-2=0 to reach that conclusion.

x+6=0

Subtract 6 on both sides:

x=-6

5x-2=0

Add 2 on both sides:

5x=2

Divide both sides by 5:

x=2/5

Since the parabola faces us and $(x+6)(5x-2)$ then we are looking at the interval from x=-6 to x=2/5 as our solution. That part is where the parabola is below the x-axis. We are looking for where it is below since it says the where is the parabola<0.

The solution in interval notation is:

$(-6,\frac{2}{5})$ .

The solution in inequality notation is:

$-6$ .

6 0

2 years ago

2.8x-1.6y=3.2 -0.7x+1.9y=6.7