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DaniilM [7]
3 years ago
12

Carl is boarding a plane. He has 2 checked bags of equal weight and a backpack that weighs 4 kg. The total weight of Carl's bagg

age is 35kg
Write an equation to determine the weight, w, of each of Carl's checked bags.
And the weight of each of Carl's check bags
Mathematics
1 answer:
g100num [7]3 years ago
3 0

Answer:

(35 - 4)/2 = w and 15.5kg

Step-by-step explanation:

To get the weight of both bags you would have to first subtract the backpack's weight from the total weight. 35 - 4 = 31. Now divide by 2 to get the weight of one bag. 31/2 = 15.5. Each bag weighs 15.5 kg

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-4(3)=-12
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Determine the area enclosed by y=2x+3, the x-axis and the ordinates x=3 and x=4​
jok3333 [9.3K]

Answer:

\displaystyle \int\limits^4_3 {2x + 3} \, dx = 10

General Formulas and Concepts:
<u>Calculus</u>

Integration

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

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:

<u>Step 1: Define</u>

<em>Identify.</em>

y = 2x + 3

<em>x</em>-interval [3, 4]

<em>x</em>-axis

<em>See attachment for graph.</em>

<u>Step 2: Find Area</u>

  1. Substitute in variables [Area of a Region Formula]:                               \displaystyle A = \int\limits^4_3 {2x + 3} \, dx
  2. [Integral] Rewrite [Integration Property - Addition/Subtraction]:           \displaystyle A = \int\limits^4_3 {2x} \, dx + \int\limits^4_3 {3} \, dx
  3. [Integrals] Rewrite [Integration Property - Multiplied Constant]:           \displaystyle A = 2 \int\limits^4_3 {x} \, dx + 3 \int\limits^4_3 {} \, dx
  4. [Integrals] Integrate [Integration Rule - Reverse Power Rule]:               \displaystyle A = 2 \bigg( \frac{x^2}{2} \bigg) \bigg| \limits^4_3 + 3(x) \bigg| \limits^4_3
  5. [Integrals] Integrate [Integration Rule - FTC 1]:                                       \displaystyle A = 2 \bigg( \frac{7}{2} \bigg) + 3(1)
  6. Simplify:                                                                                                     \displaystyle A = 10

∴ the area bounded by the region y = 2x + 3, x-axis, and the coordinates x = 3 and x = 4 is equal to 10.

---

Learn more about integration: brainly.com/question/26401241

Learn more about calculus: brainly.com/question/20197752

---

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

Unit: Integration

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