Mathematics, 18.03.2021 01:50 thibeauxkristy

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Answer from: 1162

$\displaystyle t = \frac{2 \pi}{3}, \ \frac{4 \pi}{3}$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets Parenthesis Exponents Multiplication Division Addition Subtraction Left to Right

Equality Properties

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

Algebra I

FunctionsFunction Notation[Interval Notation] - [Brackets] imply inclusive, (Parenthesis) imply exclusive

Pre-Calculus

Unit Circle

Calculus

Derivatives

Derivative Notation

The definition of a derivative is the slope of the tangent line

Basic Power Rule:

f(x) = cxⁿ f’(x) = c·nxⁿ⁻¹

Derivative Property [Addition/Subtraction]: $\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]$

Derivative Property [Multiplied Constant]: $\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)$

Trig Derivative: $\displaystyle \frac{d}{dx}[sinu] = u'cosu$

Step-by-step explanation:

Step 1: Define

[Given] q(t) = t + 2sint

[Given] Interval [0, 2π]

[Solve] q'(t) = 0

Horizontal tangent line have a slope of 0

Step 2: Differentiate

[Derivative] Basic Power Rule/Trig Derivative [Derivative Prop - Add]: $\displaystyle q'(t) = 1 \cdot t^{1 - 1} + 1 \cdot 2cost$ [Derivative] Simplify exponent: $\displaystyle q'(t) = 1 \cdot t^{0} + 1 \cdot 2cost$ [Derivative] Evaluate exponent: $\displaystyle q'(t) = 1 \cdot 1 + 1 \cdot 2cost$ [Derivative] Multiply: $\displaystyle q'(t) = 1 + 2cost$

Step 3: Solve

[Derivative] Substitute in function value: $\displaystyle 0 = 1 + 2cost$ [Subtraction Property of Equality] Isolate t term: $\displaystyle -1 = 2cost$ Rewrite: $\displaystyle 2cost = -1$ [Division Property of Equality] Isolate trig t term: $\displaystyle cost = \frac{-1}{2}$ [Equality Property] Inverse Trig: $\displaystyle t = cos^{-1}(\frac{-1}{2})$ Evaluate [Unit Circle, Interval]: $\displaystyle t = \frac{2 \pi}{3}, \ \frac{4 \pi}{3}$

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

Unit: Derivatives

Book: College Calculus 10e

Answer from: Quest

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Answer from: Quest

he used the algebra tiles correctly.

step-by-step explanation:

given : 3x(2x-1)

solution :

since he used algebra tiles .

armando's solution is the same in the solution shown in figure

as he distributed 2x into x and x and wrote them vertically

and also write -1 vertically

and 3x into x and x and x and wrote it horizontally

performing the multiplication in the algebra tiles we get $6x^{2} -3x$

and solving the given expression :

3x(2x-1)

$6x^{2} -3x$

since solving the expression using tiles and solving the expression algebraically we get the same answer

thus we can choose the first statement as the answer to this problem which is "he used the algebra tiles correctly".

Which is true regarding armando’s use of algebra tiles? he used the algebra tiles correctly. he did

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