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Helen [10]
3 years ago
5

Six more than a quotient of a number and three is greater than 14

Mathematics
2 answers:
Mashcka [7]3 years ago
6 0
Its actually really simple. x/3 + 6> 14 , or rather, x divided by 3 + 6 is greater than 14
Lorico [155]3 years ago
3 0

Remark

You sure do have to read that a couple of times.

The quotient is an answer after a division takes place So you are dividing x/3

When you have figured out that phrase you add six onto it.

(x/3) + 6 > 14 Subtract 6 from both sides.

(x/3) +6 - 6 > 14 - 6

(x/3) > 8 Now multiply by 3

x > 8*3

x > 24 <<<<<< Answer

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Step-by-step explanation:

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now do y-4 = -3/5x - 6

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hope this is right

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Step-by-step explanation:

recall in a linear equation expressed in the form

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If you compare these definitions to what was given in the question, the rate of  change in snowfall is given as 1/2 inches per hour. This is equivalent to the m value in our general equation above. Hence we can say that the rate of change in snowfall is equal to the slope of the graph.

The question also states that before the snow even started to fall (i.e when x=0 hours), there was already 8 inches on the ground. This is equivalent to the value of b in our general equation above. Hence we can say that the y-intercept is representative of the 8 inches that was already on the ground when the snow started falling.

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In 1859, 24 rabbits were released into the wild in Australia, where they had no natural predators. Their population grew exponen
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Answer:

a) P' = P

   P(t) = 24e^{0.693t} where t is step of 6 months

b) 7.7 years

c)1064.67 rabbits/year

Step-by-step explanation:

The differential equation describing the population growth is

\frac{dP}{dt} = P

Where t is the range of 6 months, or half of a year.

P(t) would have the form of

P(t) = P_0e^{kt}

where P_0 = 24 is the initial population

After 6 month (t = 1), the population is doubled to 48

P(1) = 24e^k = 48

e^k = 2

k = ln(2) = 0.693

Therefore P(t) = 24e^{0.693t}

where t is step of 6 months

b. We can solve for t to get how long it takes to get to a population of 1,000,000:

24e^{0.693t} = 1000000

e^{0.693t} = 1000000 / 24 = 41667

0.693t = ln(41667) = 10.64

t = 10.64 / 0.693 = 15.35

So it would take 15.35 * 0.5 = 7.7 years to reach 1000000

c. P' = P_0ke^{kt}

We need to resolve for k if t is in the range of 1 year. In half of a year (t = 0.5), the population is 48

24e^{0.5k) = 48

0.5k = ln2 = 0.693

k = 1.386

Therefore, P' = 1.386*24e^{1.386t}

At the mid of the 3rd year, where t = 2.5, we can calculate P'

P' = 1.386*24e^{1.386*2.5} = 1064.67 rabbits/year

4 0
3 years ago
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