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Arturiano [62]
4 years ago
9

What is the answer of this problem if mr.reyes gave his daughter P250 on her 18th birthday and intends to increase this by P150

each year.How much will her daughter receive on her 28th birthday?
Mathematics
2 answers:
Ronch [10]4 years ago
6 0
250 + (150 x 10) = 1750
o-na [289]4 years ago
4 0
28 - 18 = 10.
150 x 10 = 1500
So all you've got to do is..
150 x 1500 = ???

You can do the rest lol
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Help evaluating the indefinite integral
Dafna11 [192]

Answer:

\displaystyle \int {\frac{x}{\sqrt{4 - x^2}}} \, dx = \boxed{ -\sqrt{4 - x^2} + C }

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

Differentiation

  • Derivatives
  • Derivative Notation

Derivative Property [Multiplied Constant]:
\displaystyle (cu)' = cu'

Derivative Property [Addition/Subtraction]:
\displaystyle (u + v)' = u' + v'
Derivative Rule [Basic Power Rule]:

  1. f(x) = cxⁿ
  2. f’(x) = c·nxⁿ⁻¹

Integration

  • Integrals

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

Integration Methods: U-Substitution and U-Solve

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify given.</em>

<em />\displaystyle \int {\frac{x}{\sqrt{4 - x^2}}} \, dx

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

<em>Identify variables for u-substitution/u-solve</em>.

  1. Set <em>u</em>:
    \displaystyle u = 4 - x^2
  2. [<em>u</em>] Differentiate [Derivative Rules and Properties]:
    \displaystyle du = -2x \ dx
  3. [<em>du</em>] Rewrite [U-Solve]:
    \displaystyle dx = \frac{-1}{2x} \ du

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

  1. [Integral] Apply U-Solve:
    \displaystyle \int {\frac{x}{\sqrt{4 - x^2}}} \, dx = \int {\frac{-x}{2x\sqrt{u}}} \, du
  2. [Integrand] Simplify:
    \displaystyle \int {\frac{x}{\sqrt{4 - x^2}}} \, dx = \int {\frac{-1}{2\sqrt{u}}} \, du
  3. [Integral] Rewrite [Integration Property - Multiplied Constant]:
    \displaystyle \int {\frac{x}{\sqrt{4 - x^2}}} \, dx = \frac{-1}{2} \int {\frac{1}{\sqrt{u}}} \, du
  4. [Integral] Apply Integration Rule [Reverse Power Rule]:
    \displaystyle \int {\frac{x}{\sqrt{4 - x^2}}} \, dx = -\sqrt{u} + C
  5. [<em>u</em>] Back-substitute:
    \displaystyle \int {\frac{x}{\sqrt{4 - x^2}}} \, dx = \boxed{ -\sqrt{4 - x^2} + C }

∴ we have used u-solve (u-substitution) to <em>find</em> the indefinite integral.

---

Learn more about integration: brainly.com/question/27746495

Learn more about Calculus: brainly.com/question/27746485

---

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

Unit: Integration

5 0
2 years ago
What number should be added to 77 to make the sum 0?
labwork [276]
You should add -77 to 77 because it will cancel out.
5 0
3 years ago
Read 2 more answers
(5+5i)(5-5i)=what????
Irina18 [472]
(First,Outer,Inner,Last)
25-25i+25i-25
0
3 0
3 years ago
Read 2 more answers
(10.02)
ollegr [7]
<h2>Hello!</h2>

The answer is:

C. Cosine is negative in Quadrant III

<h2>Why?</h2>

Let's discard each given option in order to find the correct:

A. Tangent is negative in Quadrant I: It's false, all functions are positive in Quadrant I (0° to 90°).

B. Sine is negative in Quadrant II: It's false, sine is negative in positive in Quadrant II. Sine function is always positive coming from 90° to 180°.

C. Cosine is negative in Quadrant III. It's true, cosine and sine functions are negative in Quadrant III (180° to 270°), meaning that only tangent and cotangent functions will be positive in Quadrant III.

D. Sine is positive in Quadrant IV: It's false, sine is negative in Quadrant IV. Only cosine and secant functions are positive in Quadrant IV (270° to 360°)

Have a nice day!

6 0
3 years ago
Simplify the rational expression. State any restrictions on the variable. t^2-4t-32/t-8
Leno4ka [110]

1. Factor the expression t^2-4t-32:

t^2-4t-32=(t-8)(t+4).

2. Since t-8 is placed in the denominator of the given expression, then t\neq 8.

3. Now the expression can be simplified:

\dfrac{t^2-4t-32}{t-8}=\dfrac{(t-8)(t+4)}{t-8}=t+4.

5 0
4 years ago
Read 2 more answers
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