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gregori [183]
2 years ago
14

Martha is a salon owner. Yesterday, she did 5 haircuts and colored the hair of 2 clients, charging a total of $421. Today, she d

id 1 haircut and colored the hair of 2 clients, charging a total of $233. How much does Martha charge for her services? Martha charges $_______ for a haircut and $_________ for a coloring.
Mathematics
1 answer:
Vanyuwa [196]2 years ago
4 0

Answer:

  • I'm not sure if I got this correct but here's my answer. Martha charges $60.14 for a haircut and she charges $77.66 for hair coloring.
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beks73 [17]
D=dollars earned
T=time spent painting the fence.

(1,5) (2,10) (3,15)

D=5T
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3 years ago
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Solve: 3 = 27^2x+1<br><br> please solve quickly! thank you!
kap26 [50]

Answer: I believe this will be your answer for this.

<u>x = </u>2/729<u>  so basically this is your solution of x equaling to 2 over 729.</u>

<u>(hope this helps!)</u>

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2 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}

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3 years ago
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Bogdan [553]

Answer:

<h2>x = 4.8</h2>

Step-by-step explanation:

\dfrac{x}{-12}=-0.4\qquad\text{multiply both sides by (-12)}\\\\-12\!\!\!\!\!\diagup^1\cdot\dfrac{x}{-12\!\!\!\!\!\diagup_1}=(-12)(-0.4)\\\\x=4.8

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nydimaria [60]
(-2,-1)  and (1,-5) is the answer

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