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katen-ka-za [31]
3 years ago
8

Ms. Maxwell bought a purse with 1/4 of her money. The purse cost 37.00. How much money did Ms. Maxwell have left after she bough

t the purse?
Mathematics
1 answer:
bixtya [17]3 years ago
7 0

Answer:111

the purse was 1/4 so. we have to do

37.00x4 we get 148 then. we do 148-37 to get a total of 111 (:

Step-by-step explanation:

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Find the area of the region enclosed by the graphs of the functions
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Answer:

\displaystyle A = \frac{8}{21}

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

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Area - Integrals

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

Integration Rule [Fundamental Theorem of Calculus 1]:                                      \displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)

Integration Property [Addition/Subtraction]:                                                          \displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx

Area of a Region Formula:                                                                                     \displaystyle A = \int\limits^b_a {[f(x) - g(x)]} \, dx

Step-by-step explanation:

*Note:

<em>Remember that for the Area of a Region, it is top function minus bottom function.</em>

<u />

<u>Step 1: Define</u>

f(x) = x²

g(x) = x⁶

Bounded (Partitioned) by x-axis

<u>Step 2: Identify Bounds of Integration</u>

<em>Find where the functions intersect (x-values) to determine the bounds of integration.</em>

Simply graph the functions to see where the functions intersect (See Graph Attachment).

Interval: [-1, 1]

Lower bound: -1

Upper Bound: 1

<u>Step 3: Find Area of Region</u>

<em>Integration</em>

  1. Substitute in variables [Area of a Region Formula]:                                     \displaystyle A = \int\limits^1_{-1} {[x^2 - x^6]} \, dx
  2. [Area] Rewrite [Integration Property - Subtraction]:                                     \displaystyle A = \int\limits^1_{-1} {x^2} \, dx - \int\limits^1_{-1} {x^6} \, dx
  3. [Area] Integrate [Integration Rule - Reverse Power Rule]:                           \displaystyle A = \frac{x^3}{3} \bigg| \limit^1_{-1} - \frac{x^7}{7} \bigg| \limit^1_{-1}
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Topic: AP Calculus AB/BC (Calculus I/II)  

Unit: Area Under the Curve - Area of a Region (Integration)  

Book: College Calculus 10e

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