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Simora [160]
4 years ago
8

Evaluate the integral. (See picture)

Mathematics
1 answer:
Mnenie [13.5K]4 years ago
3 0
I get 25. I don't know right or wrong.
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Find the function y = f(t) passing through the point (0, 18) with the given first derivative.
monitta

Answer:

\displaystyle y = \frac{t^2}{16} + 18

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

  • Multiplication Property of Equality
  • Division Property of Equality
  • Addition Property of Equality
  • Subtraction Property of Equality

<u>Algebra I</u>

  • Functions
  • Function Notation
  • Coordinates (x, y)

<u>Calculus</u>

Derivatives

Derivative Notation

Antiderivatives - Integrals

Integration Constant C

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

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify</em>

Point (0, 18)

\displaystyle \frac{dy}{dt} = \frac{1}{8} t

<u>Step 2: Find General Solution</u>

<em>Use integration</em>

  1. [Derivative] Rewrite:                                                                                         \displaystyle dy = \frac{1}{8} t\ dt
  2. [Equality Property] Integrate both sides:                                                        \displaystyle \int dy = \int {\frac{1}{8} t} \, dt
  3. [Left Integral] Integrate [Integration Rule - Reverse Power Rule]:                 \displaystyle y = \int {\frac{1}{8} t} \, dt
  4. [Right Integral] Rewrite [Integration Property - Multiplied Constant]:           \displaystyle y = \frac{1}{8}\int {t} \, dt
  5. [Right Integral] Integrate [Integration Rule - Reverse Power Rule]:              \displaystyle y = \frac{1}{8}(\frac{t^2}{2}) + C
  6. Multiply:                                                                                                             \displaystyle y = \frac{t^2}{16} + C

<u>Step 3: Find Particular Solution</u>

  1. Substitute in point [Function]:                                                                         \displaystyle 18 = \frac{0^2}{16} + C
  2. Simplify:                                                                                                             \displaystyle 18 = 0 + C
  3. Add:                                                                                                                   \displaystyle 18 = C
  4. Rewrite:                                                                                                             \displaystyle C = 18
  5. Substitute in <em>C</em> [Function]:                                                                                \displaystyle y = \frac{t^2}{16} + 18

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

Unit: Integration

Book: College Calculus 10e

4 0
3 years ago
Rectangle J’K’L’M’ shown on the grid is the image of rectangle JKLM after transformation. The same transformation will be applie
7nadin3 [17]
The answer is C. (16, 7)

due to every x is incremented by 10

and every y is incremented by 3

the position of T is (6, 4) so the position of T' becomes (6+10, 4+3) which is (16, 7)
5 0
3 years ago
Solve for 'm': m - 7 = -13 + m
lesya [120]

Answer:

m - 7 = -13 + m

Step-by-step explanation:

8 0
4 years ago
5. Use your answers from Questions 3 and 4 to help you answer this question.
Artist 52 [7]

Answer:

^

Step-by-step explanation:

6 0
3 years ago
State the degree: 11m^3n^2p
Vadim26 [7]
<h3>Answer:  6</h3>

Explanation:

Using the rule that x = x^1, we can rewrite the p as p^1

So 11m^3n^2p is the same as 11m^3n^2p^1

The exponents are: 3, 2, 1

Those exponents add up to 3+2+1 = 6

The degree of a monomial like this is simply equal to the sum of the exponents.

7 0
3 years ago
Read 2 more answers
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