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Naddik [55]
2 years ago
10

fredy the frog was 18 feet down below ground in a well and is trying to climb out the first day he climbed up 7 feet but slid ba

ck down feet the next day he climbed 3 feet but slid backdown 4 feet write this situation out in terms of opertating with integers then find out where fredy is at now(remeber to write your answer with the correct units)
Mathematics
1 answer:
lisov135 [29]2 years ago
7 0

Writing this situation out in terms of operating with integers and indicating Fredy's position at the moment is as follows:

Fredy's position in the well = <u>13 feet in depth</u> or 5 feet from the bottom.

<h3>What is an integer?</h3>

An integer is a whole number, both positive and negative.

<h3>Data and Calculations:</h3>

Depth of well = 18 feet

Height climbed the first day = 7 feet

Height climbed the second day = 4 feet

Height of climb up the well = 11 feet

Depth of slid on the first day = 2 feet

Depth of slid on the second day = 4 feet

Total slid down the well = 6 feet (2 + 4)

The position of Fredy the Frog is at 13 feet (18 - 11 + 6) depth

Thus, based on integer operations, we can conclude that Fredy the Frog has only climbed <u>5 feet</u> (18 - 13) from the well's depth.

Learn more about integers at brainly.com/question/17695139

#SPJ1

<h3>Question Completion:</h3>

Fredy the frog was 18 feet down below ground in a well and is trying to climb out. The first day he climbed up 7 feet but slid back down <u>2 feet</u> the next day he climbed 3 feet but slid back down 4 feet.

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Lyrx [107]
How did you do that
5 0
3 years ago
A privately owned lake contains two types of game fish, bass and trout. The owner provides two types of food, A and B, for these
Vinvika [58]

Answer:

133 fishes

Step-by-step explanation:

Units of food A = 400 units

Units of food B = 400 units

Fish Bass required 2 units of A and 4 units of B.

Fish Trout requires 5 units of A and 2 units of B.

i. For food A,

total units of food A required = 2 + 5

                                                 = 7 units

number of bass and trout that would consume food A = 2 x \frac{400}{7}

                                               = 114.3

number of bass and trout that would consume food A = 114

ii. For food B,

total units of food B required = 4 + 2

                                                = 6 units

number of bass and trout that would consume food B = 2 x \frac{400}{6}

                                                 = 133.3

number of bass and trout that would consume food B = 133

Thus, the maximum number of fish that the lake can support is 133.

4 0
3 years ago
During optimal conditions, the rate of change of the population of a certain organism is proportional to the population at time
Lana71 [14]

Answer:

The population is of 500 after 10.22 hours.

Step-by-step explanation:

The rate of change of the population of a certain organism is proportional to the population at time t, in hours.

This means that the population can be modeled by the following differential equation:

\frac{dP}{dt} = Pr

In which r is the growth rate.

Solving by separation of variables, then integrating both sides, we have that:

\frac{dP}{P} = r dt

\int \frac{dP}{P} = \int r dt

\ln{P} = rt + K

Applying the exponential to both sides:

P(t) = Ke^{rt}

In which K is the initial population.

At time t = 0 hours, the population is 300.

This means that K = 300. So

P(t) = 300e^{rt}

At time t = 24 hours, the population is 1000.

This means that P(24) = 1000. We use this to find the growth rate. So

P(t) = 300e^{rt}

1000 = 300e^{24r}

e^{24r} = \frac{1000}{300}

e^{24r} = \frac{10}{3}

\ln{e^{24r}} = \ln{\frac{10}{3}}

24r = \ln{\frac{10}{3}}

r = \frac{\ln{\frac{10}{3}}}{24}

r = 0.05

So

P(t) = 300e^{0.05t}

At what time t is the population 500?

This is t for which P(t) = 500. So

P(t) = 300e^{0.05t}

500 = 300e^{0.05t}

e^{0.05t} = \frac{500}{300}

e^{0.05t} = \frac{5}{3}

\ln{e^{0.05t}} = \ln{\frac{5}{3}}

0.05t = \ln{\frac{5}{3}}

t = \frac{\ln{\frac{5}{3}}}{0.05}

t = 10.22

The population is of 500 after 10.22 hours.

7 0
3 years ago
According to the Guinness book of world records the heaviest baby ever born was 29 pounds , 4 ounces what was the baby’s mass in
viktelen [127]

Answer: 13.268 kilograms

Step-by-step explanation:

29.25 lb × 0.45359237

= 13.267576823 kg

Hope this helps!!! Good luck!!! ;)

3 0
3 years ago
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