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3 years ago

Solving for Unknown Angle Measures

Mathematics

2 answers:

Basile [38]3 years ago

6 0

Answer:

1 - 110

2 - 110

3 - 70

4 - 70

Tresset [83]3 years ago

4 0

Answer:

angle 4 is a full angle because it is represented full 110

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Easy points ^^

Shkiper50 [21]

Step-by-step explanation:

x = Alice's money

y = Mio's money

z = Grace's money

x + y + z = sum

there are originality 5+4+3 = 12 parts of the money shared between them.

x = 5/12 sum

y = 4/12 sum

z = 3/12 sum

after the money transfer we are only having 4+3+2 = 9 parts of the same money distributed.

so, we are dealing with 1/9 parts.

let's bring everything to 1/36 to be able to truly compare things.

therefore

x = 15/36 sum

y = 12/36 sum

z = 9/36 sum

15/36 sum + 10 = 16/36 sum

because 4/9 = 16/36

10 = 1/36 sum

sum = 10×36 = £360

so, together they have £360.

and they had before the £10 transfer £150, £120, £90.

after the £10 transfer they had £160, £120, £80.

5 0

2 years ago

The half-life of a certain tranquilizer in the bloodstream is 22 hours. How long will it take for the drug to decay to 90% of

nasty-shy [4]

Answer:

It will take approximately 3.34 hours for the drug to decay to 90% of the original dosage

Step-by-step explanation:

As suggested, we use the formula for exponential decay:

$A(t)=A_0\,e^{-k\,t}$

From the given information, the half life of the drug in blood id 22 hours, so that means that it takes that number of hours to go from the initial value $A_0$ , to a final value equal to $A_0/2$ . Using this information we can find the decay rate "k" by solving for this parameter in the formula, and using the natural log function to bring the exponent down:

$A(t)=A_0\,e^{-k\,t}\\\frac{A_0}{2} =A_0\,e^{-k\,*22}\\\frac{A_0}{A_0*2} =e^{-k\,*22}\\\frac{1}{2} =e^{-k\,*22}\\ln(\frac{1}{2})=-k\,*22\\ k=-\frac{ln(\frac{1}{2})}{22} \\k=0.0315$

Now we use this value for the decay rate "k" to calculate how long it would take to decay to 90% of the original dose;

$A(t)=A_0\,e^{-0.0315\,t}\\0.9*A_0} =A_0\,e^{-0.0315\,t}\\\frac{0.9*A_0}{A_0} =e^{-0.0315\,t}\\0.9 =e^{-0.0315\,t}\\ln(0.9)=-0.0315\,t\\ t=-\frac{ln(0.9)}{0.0315} \\t=3.3447\,hours$