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seropon [69]
3 years ago
9

The wholesale club sells a large box of crackers with 12 reams of crackers for $9.89. Your local store’s box of crackers has 4 r

eams for $4.29. How much do you save by buying the 12 reams from the wholesale club?
Mathematics
1 answer:
Semenov [28]3 years ago
5 0
To start off take the number of reams of crackers and divide by the total cost of the box: 12/9.89=1.21 and 4/4.29=0.93 to find out how much you pay per ream.
Then what I did was 12x0.93=11.16 (to find out how much you'd be paying with the local price)
Finally subtract 11.16-9.89= 1.27.
So in total you'd save $1.27.
Hope this helps
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Find the function y = f(t) passing through the point (0, 18) with the given first derivative.
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Answer:

\displaystyle y = \frac{t^2}{16} + 18

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<u>Algebra I</u>

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Integration Rule [Reverse Power Rule]:                                                                   \displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C

Integration Property [Multiplied Constant]:                                                             \displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify</em>

Point (0, 18)

\displaystyle \frac{dy}{dt} = \frac{1}{8} t

<u>Step 2: Find General Solution</u>

<em>Use integration</em>

  1. [Derivative] Rewrite:                                                                                         \displaystyle dy = \frac{1}{8} t\ dt
  2. [Equality Property] Integrate both sides:                                                        \displaystyle \int dy = \int {\frac{1}{8} t} \, dt
  3. [Left Integral] Integrate [Integration Rule - Reverse Power Rule]:                 \displaystyle y = \int {\frac{1}{8} t} \, dt
  4. [Right Integral] Rewrite [Integration Property - Multiplied Constant]:           \displaystyle y = \frac{1}{8}\int {t} \, dt
  5. [Right Integral] Integrate [Integration Rule - Reverse Power Rule]:              \displaystyle y = \frac{1}{8}(\frac{t^2}{2}) + C
  6. Multiply:                                                                                                             \displaystyle y = \frac{t^2}{16} + C

<u>Step 3: Find Particular Solution</u>

  1. Substitute in point [Function]:                                                                         \displaystyle 18 = \frac{0^2}{16} + C
  2. Simplify:                                                                                                             \displaystyle 18 = 0 + C
  3. Add:                                                                                                                   \displaystyle 18 = C
  4. Rewrite:                                                                                                             \displaystyle C = 18
  5. Substitute in <em>C</em> [Function]:                                                                                \displaystyle y = \frac{t^2}{16} + 18

Topic: AP Calculus AB/BC (Calculus I/II)

Unit: Integration

Book: College Calculus 10e

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<h3>How to maximize volume of a box?</h3>

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