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N76 [4]
3 years ago
5

there are 3 second grade classes at the sunshine school one has 22students and another has 20 students if there are 63 students

in the whole second grade how many are in the third class
Mathematics
2 answers:
alex41 [277]3 years ago
8 0
 there would be 21 in the last class
n200080 [17]3 years ago
7 0
u have to add 22+20 = 42 and subtract(-) 63-42 = 21 and 21 is ur answer 
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masya89 [10]

Answer:

<h3><ABC  >  <DBC.</h3>

Step-by-step explanation:

Given < DBC = < RST and we need to prove < ABC is greater than <RST.

First given statement:

< DBC = < RST

Reason: Given.

Second given statement :

<ABC = <DBC+  <ABD.

Reason: Angle addition theorem.

<em>Note: < ABC is the sum of angles <DBC and  <ABD and we have < DBC = < RST. So it's an obvious thing that the sum of angles <DBC and  <ABD is always greater than <RST.</em>

Also,  <ABC is greater than <DBC.

Therefore, correct option for third statement is :

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Step-by-step explanation:

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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>

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  • 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
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  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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