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2 years ago

11

Yo if someone helps me with this ill give u brainliest

2 answers:

hammer [34]2 years ago

5 0

Answer:

Step-by-step explanation:

a 15090

b 4255

c dairy heifers and bulls

d 9975/15090

e how many steers there are compared to bulls

Inessa05 [86]2 years ago

4 0

Answer:

a) 15,090, addition

b) 4,255, subtraction

c) bulls and beef heifers for breeding

d) 5115/15,090 or 15/90

e) How many more calves are there than steers?

3930

Step-by-step explanation:

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Anestetic [448]

Answer:

ae

Step-by-step explanation:

4 0

2 years ago

Joe added 1,500 liters of water to his pool. How many kiloliters did he add to his pool?

Dennis_Churaev [7]

Answer:

1.5 kiloleter

Step-by-step explanation:

1 kiloleter = 1000 liters

so

1500 liters = 1.5 kiloleters

pls mark brainliest if possible lol

5 0

2 years ago

Read 2 more answers

X2+8x=-1 complete the square

Sophie [7]

(X+4)^2-15

This is the answer

6 0

2 years ago

Its easy for most people i really need help what is x? 2x+17=95

UNO [17]

Answer: 39

Step-by-step explanation:

2x+17-17=95-17 so 2x=78 if you divide both sides by two you will find x=39

3 0

3 years ago

Find all of the equilibrium solutions. Enter your answer as a list of ordered pairs (R,W), where R is the number of rabbits and

zloy xaker [14]

Answer:

$(0,0)$ $(4000,0)$ and $(500,79)$

Step-by-step explanation:

Given

See attachment for complete question

Required

Determine the equilibrium solutions

We have:

$\frac{dR}{dt} = 0.09R(1 - 0.00025R) - 0.001RW$

$\frac{dW}{dt} = -0.02W + 0.00004RW$

To solve this, we first equate $\frac{dR}{dt}$ and $\frac{dW}{dt}$ to 0.

So, we have:

$0.09R(1 - 0.00025R) - 0.001RW = 0$

$-0.02W + 0.00004RW = 0$

Factor out R in $0.09R(1 - 0.00025R) - 0.001RW = 0$

$R(0.09(1 - 0.00025R) - 0.001W) = 0$

Split

$R = 0$ or $0.09(1 - 0.00025R) - 0.001W = 0$

$R = 0$ or $0.09 - 2.25 * 10^{-5}R - 0.001W = 0$

Factor out W in $-0.02W + 0.00004RW = 0$

$W(-0.02 + 0.00004R) = 0$

Split

$W = 0$ or $-0.02 + 0.00004R = 0$

Solve for R

$-0.02 + 0.00004R = 0$

$0.00004R = 0.02$

Make R the subject

$R = \frac{0.02}{0.00004}$

$R = 500$

When $R = 500$ , we have:

$0.09 - 2.25 * 10^{-5}R - 0.001W = 0$

$0.09 -2.25 * 10^{-5} * 500 - 0.001W = 0$

$0.09 -0.01125 - 0.001W = 0$

$0.07875 - 0.001W = 0$

Collect like terms

$- 0.001W = -0.07875$

Solve for W

$W = \frac{-0.07875}{ - 0.001}$

$W = 78.75$

$W \approx 79$

$(R,W) \to (500,79)$

When $W = 0$ , we have:

$0.09 - 2.25 * 10^{-5}R - 0.001W = 0$

$0.09 - 2.25 * 10^{-5}R - 0.001*0 = 0$

$0.09 - 2.25 * 10^{-5}R = 0$

Collect like terms

$- 2.25 * 10^{-5}R = -0.09$

Solve for R

$R = \frac{-0.09}{- 2.25 * 10^{-5}}$

$R = 4000$

So, we have:

$(R,W) \to (4000,0)$

When $R =0$ , we have:

$-0.02W + 0.00004RW = 0$

$-0.02W + 0.00004W*0 = 0$

$-0.02W + 0 = 0$

$-0.02W = 0$

$W=0$

So, we have:

$(R,W) \to (0,0)$

Hence, the points of equilibrium are:

$(0,0)$ $(4000,0)$ and $(500,79)$

4 0

2 years ago