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Sidana [21]
3 years ago
12

PLEASE HELP!!!!! WILL GIVE BRAINLIEST!!!

Mathematics
2 answers:
maksim [4K]3 years ago
8 0

Answer:

a. t[c(h)] = 750h

Step-by-step explanation:

The problem asks you to write a function that will help Emily calculate the number of calories she will burn per hour while biking with her sister in the bike trailer.

Normally, she burns 250 calories an hour. However, with her sister, she burns three times as much, so 3*250 = 750 calories an hour.

The total number of calories she will burn is:

a. t[c(h)] = 750h

In which h is the number of hours.

enyata [817]3 years ago
6 0

She burns 250 by herself and she burns 3 times as many with her sister.

Multiply the calories by herself by 3:

250 x 3 = 750

Now you need to multiply that by the number of hours.

The formula is a. t[c(h)] = 750h

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Answer:4 times x

4(x)

Step-by-step explanation:

4 times as many songs on Lou's music player

Lou's music player is x

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Find the slope of the line connecting the point (4,8) to the point (8,12)
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Answer: 8/8

Step-by-step explanation:

6 0
3 years ago
The pendulum swings at an arc of 30 cm and its successive swings reduced to 10% of the previous swings. What is the total distan
OleMash [197]

Answer:

S_g=  33.333cm

Step-by-step explanation:

From the question we are told that

Arc of 30 cm

10% reduction after each swing

Generally sum of geometric sequence is given by the equation

    S_g=\frac{a(1-r^n)}{1-r}

Mathematically solving for total distance traveled we have

    S_g=\frac{30(1-(0.10^5)}{1-(0.10)}

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Therefore the total distance traveled  by the pendulum after 5 swings

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3 0
3 years ago
A given field mouse population satisfies the differential equation dp dt = 0.5p − 410 where p is the number of mice and t is the
ohaa [14]

Answer:

a) t = 2 *ln(\frac{82}{5}) =5.595

b) t = 2 *ln(-\frac{820}{p_0 -820})

c) p_0 = 820-\frac{820}{e^6}

Step-by-step explanation:

For this case we have the following differential equation:

\frac{dp}{dt}=\frac{1}{2} (p-820)

And if we rewrite the expression we got:

\frac{dp}{p-820}= \frac{1}{2} dt

If we integrate both sides we have:

ln|P-820|= \frac{1}{2}t +c

Using exponential on both sides we got:

P= 820 + P_o e^{1/2t}

Part a

For this case we know that p(0) = 770 so we have this:

770 = 820 + P_o e^0

P_o = -50

So then our model would be given by:

P(t) = -50e^{1/2t} +820

And if we want to find at which time the population would be extinct we have:

0=-50 e^{1/2 t} +820

\frac{820}{50} = e^{1/2 t}

Using natural log on both sides we got:

ln(\frac{82}{5}) = \frac{1}{2}t

And solving for t we got:

t = 2 *ln(\frac{82}{5}) =5.595

Part b

For this case we know that p(0) = p0 so we have this:

p_0 = 820 + P_o e^0

P_o = p_0 -820

So then our model would be given by:

P(t) = (p_o -820)e^{1/2t} +820

And if we want to find at which time the population would be extinct we have:

0=(p_o -820)e^{1/2 t} +820

-\frac{820}{p_0 -820} = e^{1/2 t}

Using natural log on both sides we got:

ln(-\frac{820}{p_0 -820}) = \frac{1}{2}t

And solving for t we got:

t = 2 *ln(-\frac{820}{p_0 -820})

Part c

For this case we want to find the initial population if we know that the population become extinct in 1 year = 12 months. Using the equation founded on part b we got:

12 = 2 *ln(\frac{820}{820-p_0})

6 = ln (\frac{820}{820-p_0})

Using exponentials we got:

e^6 = \frac{820}{820-p_0}

(820-p_0) e^6 = 820

820-p_0 = \frac{820}{e^6}

p_0 = 820-\frac{820}{e^6}

8 0
3 years ago
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-10 and 2
-10 x 2 = -20
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7 0
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