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zvonat [6]
2 years ago
12

radioactive iodine is used to determine the health of the thyroid gland it decays according to the equation y=ae^-0.0856t where

t is in days find one fourth life of this substance
Mathematics
1 answer:
KengaRu [80]2 years ago
5 0

Answer:

16.19 days.

Step-by-step explanation:

The radioactive iodine is used to determine the health of the thyroid gland and it decays according to the equation y = a(e)^{- 0.0856t} where t is in days and a is the initial amount and y is the final amount after decay for t days.

We have to find a one-fourth life of this substance.

So, \frac{a}{4} = a(e)^{- 0.0856t}

⇒ \frac{1}{4} = 0.25 = e^{- 0.0856t}

Now, taking ln both sides we get,

ln (0.25) = - 0.0856t ln e {Since, \ln a^{b} = b \ln a}

So, t = 16.19 days. (Answer)

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One leg of a right triangle has a length of 7 m. The other side have lengths that are consecutive integers. Find these lengths
Darina [25.2K]

Answer:

6 and 5

Step-by-step explanation:

Consecutive means that it is continuous, and since the triangle largest side (7) has to be less than the additive of two other sides. Therefore 6 and 5. Hope this helps.

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Solve the absolute value equation or indicate that the equation has no solution.|2x - 3= 7Select the correct choice below and, i
Whitepunk [10]

SOLUTION:

Step 1:

In this question, we are given the following:

Step 2:

In the question, we are looking at the absolute value function which says that:

Based on this, we have that:

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1 year ago
Help please!??!!?!?
vfiekz [6]

9514 1404 393

Answer:

  a) CP = SP/1.1

  b) CP = $59.50

  c) GST = $5.95

Step-by-step explanation:

a) Divide by the coefficient of CP.

  SP = 1.1×CP

  CP = SP/1.1

__

b) Use the formula with the given value.

  CP = $65.45/1.1 = $59.50

__

c) You can do this two ways: subtract CP from SP, or multiply CP by 0.1.

  GST = SP -CP = $65.45 -59.50 = $5.95

  GST = CP×0.10 = $59.50 × 0.10 = $5.95

5 0
3 years ago
What is the solution for the equation StartFraction 5 Over 3 b cubed minus 2 b squared minus 5 EndFraction = StartFraction 2 Ove
wolverine [178]

Answer:

The solutions are:

b=0,\:b=4

Step-by-step explanation:

Considering the expression

  • \frac{5}{3b^3-2b^2-5}=\frac{2}{b^3-2}

Solving the expression

\frac{5}{3b^3-2b^2-5}=\frac{2}{b^3-2}

\mathrm{Apply\:fraction\:cross\:multiply:\:if\:}\frac{a}{b}=\frac{c}{d}\mathrm{\:then\:}a\cdot \:d=b\cdot \:c

5\left(b^3-2\right)=\left(3b^3-2b^2-5\right)\cdot \:2

5b^3-10=6b^3-4b^2-10

\mathrm{Switch\:sides}

6b^3-4b^2-10=5b^3-10

6b^3-4b^2-10+10=5b^3-10+10

6b^3-4b^2=5b^3

\mathrm{Subtract\:}5b^3\mathrm{\:from\:both\:sides}

6b^3-4b^2-5b^3=5b^3-5b^3

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Using\:the\:Zero\:Factor\:Principle: if\:\mathrm ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)

So,

b=0,b-4=0

b=0,b=4

Therefore, the solutions are:

b=0,\:b=4

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Answer:

I don't really know so sorry about that.

Step-by-step explanation:

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