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andrezito [222]
3 years ago
9

Dale mixed 5.9 grams of salt into a pot of soup he was cooking. Before he served the soup, Dale added 0.31 grams of salt. How mu

ch salt did Dale put into the soup in all? EXPLAIN what you need to do, and solve it. QUIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIICKKKKKKKKKKKKKKKKKKKKK
Mathematics
1 answer:
belka [17]3 years ago
8 0

Answer:

<em>Dale put 6.21 grams of salt into the soup in all</em>

Step-by-step explanation:

<u>Arithmetic</u>

There are situations where different mathematic operations must be done in order to find the required results. Four operations are known as basics in arithmetic: addition, subtraction, multiplication, and division.

The question describes how Dale added 5.0 grams of salt into a pot of soup and then he added 0.31 grams of salt more. This means the total amount of salt added by Dale into the soup is the sum of both quantities:

Total salt added = 5.9 grams + 0.31 grams = 6.21 grams.

Dale put 6.21 grams of salt into the soup in all

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The equation for exponential growth is: P=P_oe^{kt}

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Use the half life information to find k:

\dfrac{1}{2}P_o=P_oe^{k(800)}\\\\\\\dfrac{1}{2}=e^{800k}\qquad \rightarrow \qquad \text{divided both sides by}\ P_o\\\\\\ln\bigg(\dfrac{1}{2}\bigg)=800k\qquad \rightarrow \qquad \text{applied ln to both sides}\\\\\\\dfrac{ln(\frac{1}{2})}{800}=k\qquad \rightarrow \qquad \text{divided both sides by 800}\\\\\\\large\boxed{-0.000867=k}

Next, input the information (65% decayed = 35% remaining) and the k-value to find your answer.

.35P_o=P_oe^{-0.000867t}\\\\\\.35=e^{-0.000867t}\\\\\\ln(.35)=-0.000867t\\\\\\\dfrac{ln(.35)}{-0.000867}=t\\\\\\\large\boxed{1211.6585=t}

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