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ale4655 [162]
3 years ago
10

PLEASE HELP SOLVE THESE!! DUE IN AN HOUR!! THANK YOU:))

Mathematics
1 answer:
astra-53 [7]3 years ago
7 0

Answer:

1, (2,5)

2, (6,2)

3, (-5,9)

Step-by-step explanation:

OMG I was doing the same thing solving by elimination. can I get the brainlest? Pls

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Answer:

so the answer is: eijieljnqwsmiqwmx

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Step-by-step explanation:

Destroy your keyboard by smashing each random word on the poor broken keyboard  

6 0
3 years ago
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These data show the age in years of 9 members of the youth orchestra. 12 14 16 17.5 10 18 15 15.5 19 What is the IQR (interquart
ryzh [129]
The interquartile range is 4.75.
8 0
4 years ago
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Can someone please help! Thank you so much!
torisob [31]

Answer:

a. 3x (150 degrees)

b. 9x (180 degrees)

c. 3x - 20 (220 degrees)

3 0
3 years ago
An important tool in archeological research is radiocarbon dating, developed by the American chemist Willard F. Libby.3 This is
Radda [10]

Answer:

Q(t) = Q_o*e^(-0.000120968*t)

Step-by-step explanation:

Given:

- The ODE of the life of Carbon-14:

                                       Q' = -r*Q

- The initial conditions Q(0) = Q_o

- Carbon isotope reaches its half life in t = 5730 yrs

Find:

The expression for Q(t).

Solution:

- Assuming Q(t) satisfies:

                                       Q' = -r*Q

- Separate variables:

                                      dQ / Q = -r .dt

- Integrate both sides:

                                       Ln(Q) = -r*t + C

- Make the relation for Q:

                                       Q = C*e^(-r*t)

- Using initial conditions given:

                                       Q(0) = Q_o

                                       Q_o = C*e^(-r*0)

                                      C = Q_o    

- The relation is:

                                       Q(t) = Q_o*e^(-r*t)

- We are also given that the half life of carbon is t = 5730 years:

                                       Q_o / 2 = Q_o*e^(-5730*r)

                                        -Ln(0.5) = 5730*r

                                        r = -Ln(0.5)/5730

                                        r = 0.000120968          

- Hence, our expression for Q(t) would be:

                                       Q(t) = Q_o*e^(-0.000120968*t)                                    

7 0
3 years ago
What is the area of this figure 3,7,7
IRINA_888 [86]
I think the answer is 147
4 0
3 years ago
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