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kiruha [24]
3 years ago
8

What equation equals 201?

Mathematics
2 answers:
Yanka [14]3 years ago
7 0
201=201 *1
201= 67 * 3

Hope this helps :)
Scilla [17]3 years ago
5 0
201 x 1 = 201
That is an equation that equals 201<span />
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How many cents is equal $¼​
Hitman42 [59]

Answer:

25 Cents

Step-by-step explanation:

1 / 4 = 0.25 = 25 cents

3 0
3 years ago
Factor the expression completely -15t-6
serg [7]

Answer:

\large\boxed{-15t-6=-3(5t+2)}

Step-by-step explanation:

-15t-6=(-3)(5t)+(-3)(2)=-3(5t+2)

6 0
3 years ago
The two triangles are similar. Given that a = 7, b = 17, and c = 23, find the value of x. Round your answer to the nearest hundr
Nezavi [6.7K]
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3 0
2 years ago
Showing ALL calculations, whether this statement is valid
choli [55]

Answer:

<em>R</em><em>M</em>79 608.51

Step-by-step explanation:

First year,

<em>RM</em>505 050.00×5%=<em>RM</em>505 050.00×5/100

=<em>RM</em>25 252.50

second year,

(<em>RM</em>505 050.00+<em>R</em><em>M</em>25 252.50)×5%=<em>RM</em>530 302.50×5/100

=<em>RM</em>26 515.13

third year,

(<em>R</em><em>M</em>530 302.50+<em>R</em><em>M</em>26 515.13)×5%=<em>R</em><em>M</em>556817.63×5/100

=<em>R</em><em>M</em>27840.88

three years interest,

<em>RM</em>25 252.50+<em>RM</em>26 515.13+<em>RM</em>27840.88=<em>RM</em>79608.51

<em><u>I</u></em><em><u> </u></em><em><u>d</u></em><em><u>o</u></em><em><u>n</u></em><em><u>'</u></em><em><u>t</u></em><em><u> </u></em><em><u>k</u></em><em><u>n</u></em><em><u>o</u></em><em><u>w</u></em><em><u> </u></em><em><u>t</u></em><em><u>h</u></em><em><u>a</u></em><em><u>t</u></em><em><u> </u></em><em><u>i</u></em><em><u>s</u></em><em><u> </u></em><em><u>R</u></em><em><u> </u></em><em><u>o</u></em><em><u>r</u></em><em><u> </u></em><em><u>R</u></em><em><u>M</u></em><em><u> </u></em>

<em><u>i</u></em><em><u>f</u></em><em><u> </u></em><em><u>i</u></em><em><u>t</u></em><em><u> </u></em><em><u>i</u></em><em><u>s</u></em><em><u> </u></em><em><u>R</u></em><em><u> </u></em><em><u>,</u></em><em><u>s</u></em><em><u>o</u></em><em><u>r</u></em><em><u>r</u></em><em><u>y</u></em>

<u>H</u><u>o</u><u>p</u><u>e</u><u> </u><u>c</u><u>a</u><u>n</u><u> </u><u>h</u><u>e</u><u>l</u><u>p</u><u>!</u>

8 0
3 years ago
<img src="https://tex.z-dn.net/?f=%5Cint%5Climits%5Ea_b%20%7B%281-x%5E%7B2%7D%20%29%5E%7B3%2F2%7D%20%7D%20%5C%2C%20dx" id="TexFo
Ludmilka [50]

Answer:\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{3arcsin(a) + 2a(1 - a^2)^\Big{\frac{3}{2}} + 3a\sqrt{1 - a^2}}{8} - \frac{3arcsin(b) + 2b(1 - b^2)^\Big{\frac{3}{2}} + 3b\sqrt{1 - b^2}}{8}General Formulas and Concepts:

<u>Pre-Calculus</u>

  • Trigonometric Identities

<u>Calculus</u>

Differentiation

  • Derivatives
  • Derivative Notation

Integration

  • Integrals
  • Definite/Indefinite Integrals
  • Integration Constant C

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)

U-Substitution

  • Trigonometric Substitution

Reduction Formula:                                                                                               \displaystyle \int {cos^n(x)} \, dx = \frac{n - 1}{n}\int {cos^{n - 2}(x)} \, dx + \frac{cos^{n - 1}(x)sin(x)}{n}

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify</em>

\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx

<u>Step 2: Integrate Pt. 1</u>

<em>Identify variables for u-substitution (trigonometric substitution).</em>

  1. Set <em>u</em>:                                                                                                             \displaystyle x = sin(u)
  2. [<em>u</em>] Differentiate [Trigonometric Differentiation]:                                         \displaystyle dx = cos(u) \ du
  3. Rewrite <em>u</em>:                                                                                                       \displaystyle u = arcsin(x)

<u>Step 3: Integrate Pt. 2</u>

  1. [Integral] Trigonometric Substitution:                                                           \displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \int\limits^a_b {cos(u)[1 - sin^2(u)]^\Big{\frac{3}{2}} \, du
  2. [Integrand] Rewrite:                                                                                       \displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \int\limits^a_b {cos(u)[cos^2(u)]^\Big{\frac{3}{2}} \, du
  3. [Integrand] Simplify:                                                                                       \displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \int\limits^a_b {cos^4(u)} \, du
  4. [Integral] Reduction Formula:                                                                       \displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{4 - 1}{4}\int \limits^a_b {cos^{4 - 2}(x)} \, dx + \frac{cos^{4 - 1}(u)sin(u)}{4} \bigg| \limits^a_b
  5. [Integral] Simplify:                                                                                         \displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{cos^3(u)sin(u)}{4} \bigg| \limits^a_b + \frac{3}{4}\int\limits^a_b {cos^2(u)} \, du
  6. [Integral] Reduction Formula:                                                                          \displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{cos^3(u)sin(u)}{4} \bigg|\limits^a_b + \frac{3}{4} \bigg[ \frac{2 - 1}{2}\int\limits^a_b {cos^{2 - 2}(u)} \, du + \frac{cos^{2 - 1}(u)sin(u)}{2} \bigg| \limits^a_b \bigg]
  7. [Integral] Simplify:                                                                                         \displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{cos^3(u)sin(u)}{4} \bigg| \limits^a_b + \frac{3}{4} \bigg[ \frac{1}{2}\int\limits^a_b {} \, du + \frac{cos(u)sin(u)}{2} \bigg| \limits^a_b \bigg]
  8. [Integral] Reverse Power Rule:                                                                     \displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{cos^3(u)sin(u)}{4} \bigg| \limits^a_b + \frac{3}{4} \bigg[ \frac{1}{2}(u) \bigg| \limits^a_b + \frac{cos(u)sin(u)}{2} \bigg| \limits^a_b \bigg]
  9. Simplify:                                                                                                         \displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{cos^3(u)sin(u)}{4} \bigg| \limits^a_b + \frac{3cos(u)sin(u)}{8} \bigg| \limits^a_b + \frac{3}{8}(u) \bigg| \limits^a_b
  10. Back-Substitute:                                                                                               \displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{cos^3(arcsin(x))sin(arcsin(x))}{4} \bigg| \limits^a_b + \frac{3cos(arcsin(x))sin(arcsin(x))}{8} \bigg| \limits^a_b + \frac{3}{8}(arcsin(x)) \bigg| \limits^a_b
  11. Simplify:                                                                                                         \displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{3arcsin(x)}{8} \bigg| \limits^a_b + \frac{x(1 - x^2)^\Big{\frac{3}{2}}}{4} \bigg| \limits^a_b + \frac{3x\sqrt{1 - x^2}}{8} \bigg| \limits^a_b
  12. Rewrite:                                                                                                         \displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{3arcsin(x) + 2x(1 - x^2)^\Big{\frac{3}{2}} + 3x\sqrt{1 - x^2}}{8} \bigg| \limits^a_b
  13. Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]:              \displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{3arcsin(a) + 2a(1 - a^2)^\Big{\frac{3}{2}} + 3a\sqrt{1 - a^2}}{8} - \frac{3arcsin(b) + 2b(1 - b^2)^\Big{\frac{3}{2}} + 3b\sqrt{1 - b^2}}{8}

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

Unit: Integration

Book: College Calculus 10e

8 0
2 years ago
Read 2 more answers
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