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vodomira [7]
1 year ago
11

Molly is studying two number patterns. Pattern A starts at 0 and has the rule "add 3." Pattern B starts at 0 and has the rule "a

dd 6." What is the relationship between corresponding terms in the two patterns?
The terms in Pattern A are 3 times the corresponding terms in Pattern B.
The terms in Pattern A are 2 times the corresponding terms in Pattern B.
The terms in Pattern A are one half the corresponding terms in Pattern B.
The terms in Pattern A are one third the corresponding terms in Pattern B.
Mathematics
1 answer:
tekilochka [14]1 year ago
6 0

The relation between the Pattern A is the half of the Pattern B is given below. Then the correct option is C.

<h3>What are ratio and proportion?</h3>

A ratio is a collection of ordered integers a and b represented as a/b, with b never equaling zero. A proportionate expression is one in which two items are equal.

Molly is studying two number patterns.

Pattern A starts at 0 and has the rule “add 3”.

Pattern B starts at 0 and has the rule “add 6”.

The relation between the Pattern A and Pattern B is given below.

Pattern A = 1/2 x Pattern B

Then the correct option is C.

More about the ratio and the proportion link is given below.

brainly.com/question/14335762

#SPJ1

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5.5 days (nearest tenth)

Step-by-step explanation:

<u>Given formula:</u>

\sf N=N_0e^{-kt}

  • \sf N_0 = initial mass (at time t=0)
  • N = mass (at time t)
  • k = a positive constant
  • t = time (in days)

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  • \sf N_0 = 11 g
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Half-life:  The <u>time</u> required for a quantity to reduce to <u>half of its initial value</u>.

To find the time it takes (in days) for the substance to reduce to half of its initial value, substitute the given values into the formula and set N to half of the initial mass, then solve for t:

\begin{aligned}\sf N & = \sf N_0e^{-kt}\\\\\implies \sf \dfrac{11}{2} & = \sf 11e^{-0.125t}\\\\\sf \dfrac{1}{2} & = \sf e^{-0.125t}\\\\\sf \ln \left(\dfrac{1}{2}\right) & = \sf \ln e^{-0.125t\\\\\sf \ln \left(\dfrac{1}{2}\right) & = \sf -0.125t \ln e\\\\\sf \ln \left(\dfrac{1}{2}\right) & = \sf -0.125t(1)\\\sf t & = \sf \dfrac{\ln \left(\dfrac{1}{2}\right)}{-0.125}\\\\\sf t & = \sf 5.545177444...\\\\\sf t & = \sf 5.5\:\:days\:\:(nearest\:tenth)\end{aligned}

Therefore, the substance's half-life is 5.5 days (nearest tenth).

Learn more about solving exponential equations here:

brainly.com/question/28016999

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