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3 years ago

Please help me on this

1 answer:

5 0

A. 45/100 reduced = 9/20
(divided 45 and 100 by 5 to get 9 and 20)

B. she has finished 27 problems, which is 45% of her homework. In other words
45% of the total problems = 27
if we use X to represent the total amount of problems, then:
45% X = 27
we solved in part A for 45%, which is 9/20, so then:
9/20 x = 27
now all that's left is to solve for x
first multiply both sides by 20
9x = 540
now divide both sides by 9
x = 60
the total amount of problems, or the number of problems she was assigned is 60 problems.

C. If the total amount of problems she had to do was 60, and she has already done 27, then
60 - 27 = the amount of problems she has left
33 = the amount of problems she has left

she has 33 problems left to do.

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Is this statement always, sometimes, or never true.

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Answer:

Never true

Step-by-step explanation:

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3 years ago

Thomas is 4 years older than his brother william. in five years the sum of their age will be 24. how old is each boy?

Rasek [7]

Let William be x years old.

Thomas = (4 + x) years

In five years,

William = (x + 5) years

Thomas = 4 + x + 5 = (9 + x) years

Given, in five years sum of their ages is 24.

x + 5 + 9 + x = 24

2x + 14 = 24

2x = 24 - 14

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4 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 $(500,79)$

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3 years ago

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4 years ago

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Step-by-step explanation:

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3 years ago

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