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Dafna11 [192]
3 years ago
11

Three equals parentheses five squared 1/2 parentheses you parentheses X squared 1/2

Mathematics
1 answer:
Aliun [14]3 years ago
3 0

Answer:

Can you give me a picture of the question? Or be more specific?

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I don’t understand this and it’s due at 12
Ivahew [28]

Answer:

The 20% tip would be 18%

The 80% mark down would be 0.8 I think

The 60% I'm not sure for the 60% one

The 10% I'm not sure for that one but I think it would be 3.75

I think for 75% mark down it would be 0.7 I'm not sure tho and I'm not sure for the rest

Sorry if I'm wrong but hope this helps:)

4 0
3 years ago
50 POINTS!!!! HELPPPPPPPPPPPPPPPPP!!!!!!!!!!!!!!!!!!!!! WILL GIVE BRANLEST!!!
sineoko [7]
  1. B
  2. y=10 is a function, x=-3 is not a function, 3x+2x=3 is not a function, first graph (the curved one) is a function, (-2, 6), (-3, 7), (-4, 8), (-3, 10) is not a function, second graph is a function
  3. first one is a function, second is not, third is, fourth isnt
  4. (-5,2) and (-5,1), (6,-5) and (6, 5), (2,2) and (2,3), (-1, 0) and (-1, 6)

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7 0
3 years ago
Read 2 more answers
Simplify the expression: 3√32/8
gtnhenbr [62]

Answer:

48 im so sorry if its wrong ps i hope this helps

Step-by-step explanation:

4 0
4 years ago
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PLS HELP I BEG U!!!!!!!!!!!!
Reika [66]

Answer: Number six should be 14/18.

Step-by-step explanation:

The simplified answer is 7/9.

4 0
3 years ago
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Drug Reaction The intensity of the reaction to a certain drug, in appropriate units, is given by
Gemiola [76]

Answer:

a) A=0.86

b) A=2.81

c) A=2.88

Step-by-step explanation:

The average of a functions can be written as:

A=\frac{1}{b-a}\int^{b}_{a}f(t)dt (1)

a.) The second hour means that the interval of time is from 0 to 2 hours, so a=0 and b=2. Using the equation (1) we can calculate the average intensity.

A=\frac{1}{2-0}\int^{2}_{0}te^{-0.1t}dt

Using integration by parts we can solve it.

\int fdg=fg-\int gdf

f=t , df=dt

dg=e^{-t/10}dt , g=-10e^{-t/10}dt

\int^{2}_{0}te^{-0.1t}dt=-10e^{-t/10}t|^{2}_{0}+10\int^{2}_{0} e^{-t/10}dt=-10e^{-t/10}t|^{2}_{0}-100e^{-t/10}|^{2}_{0}=1.75 (2)

Therefore, the average intensity at the second hour will be:

A=\frac{1}{2-0}\int^{2}_{0}te^{-0.1t}dt=\frac{1.75}{2}=0.86

b) We can use the equation (2) to solve it. In this case the limits of integration will be a = 0 and b = 12 hours.  

\int^{12}_{0}te^{-0.1t}dt=-10e^{-t/10}t|^{12}_{0}-100e^{-t/10}|^{12}_{0}=33.74

Therefore, the average intensity at the twelfth hour will be:

A=\frac{1}{12-0}\int^{12}_{0}te^{-0.1t}dt=\frac{33.74}{12}=2.81

c) Finally, here a = 0 and b = 24 hour.

\int^{24}_{0}te^{-0.1t}dt=-10e^{-t/10}t|^{24}_{0}-100e^{-t/10}|^{12}_{0}=69.16

Therefore, the average intensity at the twenty-fourth hour will be:

A=\frac{1}{24-0}\int^{24}_{0}te^{-0.1t}dt=\frac{69.16}{24}=2.88

I hope it helps you!

7 0
4 years ago
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