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Vanyuwa [196]
3 years ago
6

Help please will give brainly

Mathematics
1 answer:
Gre4nikov [31]3 years ago
7 0

Answer:

So what it's saying is that the more hours she babysits, the amount of money she earns goes up by $5.

Since the picture doesn't show a graph, I have no idea what the real answer is.

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After a car is purchased, the value of a new car decreases $4000 each year. After 3 years, the car is worth $18000.
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Find the area of the region enclosed by the graphs of these equations. (CALCULUS HELP)
sergiy2304 [10]

Answer:

\displaystyle A = \frac{20\sqrt{15}}{3}

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

Equality Properties

  1. Multiplication Property of Equality
  2. Division Property of Equality
  3. Addition Property of Equality
  4. Subtraction Property of Equality

<u>Algebra I</u>

  • Terms/Coefficients
  • Graphing
  • Exponential Rule [Root Rewrite]:                                                                   \displaystyle \sqrt[n]{x} = x^{\frac{1}{n}}

<u>Calculus</u>

Derivatives

Derivative Notation

Derivative of a constant is 0

Basic Power Rule:

  • f(x) = cxⁿ
  • f’(x) = c·nxⁿ⁻¹

Area - Integrals

U-Substitution

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

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

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>

F: y = √(15 - x)

G: y = √(15 - 3x)

H: y = 0

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

<em>Solve each equation for the x-value for our bounds of integration.</em>

F

  1. Set <em>y</em> = 0:                                                                                                         0 = √(15 - x)
  2. [Equality Property] Square both sides:                                                          0 = 15 - x
  3. [Subtraction Property of Equality] Isolate <em>x</em> term:                                         -x = -15
  4. [Division Property of Equality] Isolate <em>x</em>:                                                        x = 15

G

  1. Set y = 0:                                                                                                         0 = √(15 - 3x)
  2. [Equality Property] Square both sides:                                                          0 = 15 - 3x
  3. [Subtraction Property of Equality] Isolate <em>x</em> term:                                         -3x = -15
  4. [Division Property of Equality] Isolate <em>x</em>:                                                        x = 5

This tells us that our bounds of integration for function F is from 0 to 15 and our bounds of integration for function G is 0 to 5.

We see that we need to subtract function G from function F to get our area of the region (See attachment graph for visual).

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

<em>Integration Part 1</em>

  1. Rewrite Area of Region Formula [Integration Property - Subtraction]:     \displaystyle A = \int\limits^b_a {f(x)} \, dx - \int\limits^d_c {g(x)} \, dx
  2. [Integral] Substitute in variables and limits [Area of Region Formula]:     \displaystyle A = \int\limits^{15}_0 {\sqrt{15 - x}} \, dx - \int\limits^5_0 {\sqrt{15 - 3x}} \, dx
  3. [Area] [Integral] Rewrite [Exponential Rule - Root Rewrite]:                       \displaystyle A = \int\limits^{15}_0 {(15 - x)^{\frac{1}{2}}} \, dx - \int\limits^5_0 {(15 - 3x)^{\frac{1}{2}}} \, dx

<u>Step 4: Identify Variables</u>

<em>Set variables for u-substitution for both integrals.</em>

Integral 1:

u = 15 - x

du = -dx

Integral 2:

z = 15 - 3x

dz = -3dx

<u>Step 5: Find Area of Region</u>

<em>Integration Part 2</em>

  1. [Area] Rewrite [Integration Property - Multiplied Constant]:                       \displaystyle A = -\int\limits^{15}_0 {-(15 - x)^{\frac{1}{2}}} \, dx + \frac{1}{3}\int\limits^5_0 {-3(15 - 3x)^{\frac{1}{2}}} \, dx
  2. [Area] U-Substitution:                                                                                   \displaystyle A = -\int\limits^0_{15} {u^{\frac{1}{2}}} \, du + \frac{1}{3}\int\limits^0_{15} {z^{\frac{1}{2}}} \, dz
  3. [Area] Reverse Power Rule:                                                                         \displaystyle A = -(\frac{2u^{\frac{3}{2}}}{3}) \bigg|\limit^0_{15} + \frac{1}{3}(\frac{2z^{\frac{3}{2}}}{3}) \bigg|\limit^0_{15}
  4. [Area] Evaluate [Integration Rule - FTC 1]:                                                   \displaystyle A = -(-10\sqrt{15}) + \frac{1}{3}(-10\sqrt{15})
  5. [Area] Multiply:                                                                                               \displaystyle A = 10\sqrt{15} + \frac{-10\sqrt{15}}{3}
  6. [Area] Add:                                                                                                     \displaystyle A = \frac{20\sqrt{15}}{3}

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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3 years ago
What is 4/10 in simplest form
Vedmedyk [2.9K]
\frac{4}{10} in simplest form would be \frac{2}{5} because you could divideboth numerator and denominator by 2. Hope it helps!!
5 0
4 years ago
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