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3 years ago

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Mathematics

1 answer:

Alecsey [184]3 years ago

4 0

Answer:

Explanation:

Since 30 people have already voted, they need 30 more to vote so that they will have 60 votes. Since Devon got 4 votes the first time we will double that to go to 60 votes which than gives you your solution of 8 answer choice B.

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URGENT PLEASE HELP Picture attached

solong [7]

Answer:

A) ∠AOB = 36°

Step-by-step explanation:

The full circle area = 10²π or 100π

the shaded portion is (10π/100π)(360°) = 36°

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3 years ago

Question 20 Only, Please show your work.<br><br> Solve the problem.

Sveta_85 [38]

Answer:

7 inches

Step-by-step explanation:

The area of a square is given by the formula ...

A = s² . . . . . s = side length

Using the given area, you have ...

49 in² = s² . . . . . fill in given value of area

7 in = s . . . . . . . take the positive square root of both sides

The length of a side of the square is 7 inches.

8 0

3 years ago

Find the difference of 42 3/4 - 34 1/8. <br> Pleasee help asap

ValentinkaMS [17]

8 5/8 or 8.625; have a good one

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3 years ago

HURRY!!!!! For the given central angle, determine the distance traveled along the unit circle from the point (1, 0). 16 degrees

Law Incorporation [45]

Answer:

c. (-0.82, -0.57)

Step-by-step explanation:

6 0

3 years ago

Radioactive Decay

SpyIntel [72]

Answer:

Percentage of (226Ra) after 900 years is 68%

Step-by-step explanation:

Let P(t) be the amount of (226Ra) present at any time t

Half life of (226 Ra) = 1599 years

If P₀ is initial amount of (226 Ra) then after 1599 years

P(1599)=P₀/2

Decay i amount of radioactive substance is related to time t as

$\frac{dP}{dt}=kP(t)\\\\\frac{1}{P}\,dP=kdt\\\\Integrating\,\, both\,\,sides\\\\ln|P|=kt+c\\\\P(t)=Ce^{kt}\\\\at \,\, t=0\,\, P(0)=P_{o}\\\\P(0)=Ce^{k0}\\\\P_{o}=C\\\\then\\\\P(t)=P_{o}e^{kt}$

To find value of k

$at\,\, t=1599\,years\\\\P(1599)=\frac{P_{o}}{2}\\\\then\\\\\frac{P_{o}}{2} =P_{o}e^{k(1599)}\\\\\frac{1}{2} =e^{k(1599)}\\\\ln|\frac{1}{2}|=k(1599)\\\\k=\frac{ln|\frac{1}{2}|}{1599}=-4.3\times 10^{-3}\\\\\implies P(t)=P_{o}e^{-4.3\times 10^{-3}t}\\\\at\,\, t=900 \\\\P(900)=P_{o}e^{-4.3\times 10^{-3}(900)}\\\\P(900)=0.68P_{o}$

Percentage of radioactive element is:

Amount after 900 years $=\frac{P(900)}{P_{o}}\times 100\\\\=\frac{0.68P_{o}}{P_{o}}\times 100\%\\\\=68\%$

3 0

3 years ago