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Gala2k [10]
3 years ago
12

I need help with this equation -17 = -7c + 4

Mathematics
2 answers:
dimulka [17.4K]3 years ago
6 0

Answer:

C = 3

Step-by-step explanation:

KonstantinChe [14]3 years ago
3 0

Answer: c = 3

Step-by-step explanation:

Hey there! I will give the following steps, if you have any questions feel free to ask me in the comments below.

<u>Step 1: </u><u><em>Subtract 4 from both sides.</em></u>

-17 - 4 = -7c

<u>Step 2:</u><u><em> Simplify -17 - 4 to -21.</em></u>

-21 = 7c

<u>Step 3:</u><u> </u><u><em>Divide both sides by -7.</em></u>

\frac{-21}{-7} = c

<u>Step 4:</u><u> </u><u><em>Two negatives make a positive.</em></u>

\frac{21}{7} = c

<u>Step 5: </u><u><em>Simplify </em></u>\frac{21}{7}<u><em> to 3.</em></u>

3 = c

<u>Step 6:</u><u> </u><u><em>Switch sides.</em></u>

c = 3

~I hope I helped you! :)~

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Answer:

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)

Given values:

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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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It may be convenient here to write the equation of the line as

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Using the first two points, we have

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You can work directly with the values in the table. The y-value increases by 5 from one line to another, while the x-value increases by 6. Thus, to make the y-value decrease by 5 from the first point, we need to decrease that point's x-value by 6 to -21. That is, the x-intercept is (-21, 0).

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