Plz help me with this quick 30 points plz give me the right answers plz help me thanks

Mathematics

1 answer:

algol [13]3 years ago

6 0

Answer:

1b. 2/3

2b. 4/3

3b. 3

Step-by-step explanation:

To get the answers to each of these, first change the percentage to a decimal. Then change the mixed number to an improper fraction. Then multiply.

1b.

50% * 1 1/3

5/10 * 4/3

20/30

2/3

2b.

40% * 3 1/3

4/10 * 10/3

40/30

4/3

3b.

80% * 3 3/4

8/10 * 15/4

120/40

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Find the value of k if it is known that the graph of y=kx goes through:

DENIUS [597]

In y=kx , subst. -30 for x and 3 for y. Then solve the resulting equation for k:

3 = k(-30).

Dividing both sides by -30, we get:

-3/30 = -1/10 = k

The constant of proportionality is k = -1/10.

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4 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

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