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- https://math.libretexts.org/Courses/Coastline_College/Math_C170%3A_Precalculus_(Tran)/12%3A_Introduction_to_Calculus/12.05%3A_DerivativesChange divided by time is one example of a rate. The rates of change in the previous examples are each different. In other words, some changed faster than others. If we were to graph the functions, we...Change divided by time is one example of a rate. The rates of change in the previous examples are each different. In other words, some changed faster than others. If we were to graph the functions, we could compare the rates by determining the slopes of the graphs.
- https://math.libretexts.org/Courses/Coastline_College/Math_C170%3A_Precalculus_(Tran)/12%3A_Introduction_to_CalculusCalculus is the broad area of mathematics dealing with such topics as instantaneous rates of change, areas under curves, and sequences and series. Underlying all of these topics is the concept of a li...Calculus is the broad area of mathematics dealing with such topics as instantaneous rates of change, areas under curves, and sequences and series. Underlying all of these topics is the concept of a limit, which consists of analyzing the behavior of a function at points ever closer to a particular point, but without ever actually reaching that point. Calculus has two basic applications: differential calculus and integral calculus.
- https://math.libretexts.org/Courses/Coastline_College/Math_C170%3A_Precalculus_(Tran)/09%3A_Systems_of_Equations_and_Inequalities/9.05%3A_Partial_FractionsDecompose a ratio of polynomials by writing the partial fractions. Solve by clearing the fractions, expanding the right side, collecting like terms, and setting corresponding coefficients equal to eac...Decompose a ratio of polynomials by writing the partial fractions. Solve by clearing the fractions, expanding the right side, collecting like terms, and setting corresponding coefficients equal to each other, then setting up and solving a system of equations. The decomposition with repeated linear factors must account for the factors of the denominator in increasing powers. The decomposition with a nonrepeated irreducible quadratic factor needs a linear numerator over the quadratic factor.
- https://math.libretexts.org/Courses/Coastline_College/Math_C170%3A_Precalculus_(Tran)/08%3A_Further_Applications_of_Trigonometry/8.E%3A_Further_Applications_of_Trigonometry_(Exercises)If the angle of elevation from the man to the balloon is \(27^{\circ}\), and the angle of elevation from the woman to the balloon is \(41^{\circ}\), find the altitude of the balloon to the nearest foo...If the angle of elevation from the man to the balloon is \(27^{\circ}\), and the angle of elevation from the woman to the balloon is \(41^{\circ}\), find the altitude of the balloon to the nearest foot. For polar coordinates, the point in the plane depends on the angle from the positive \(x\)-axis and distance from the origin, while in Cartesian coordinates, the point represents the horizontal and vertical distances from the origin.
- https://math.libretexts.org/Courses/Coastline_College/Math_C170%3A_Precalculus_(Tran)/zz%3A_Back_Matter/10%3A_Index
- https://math.libretexts.org/Courses/Fullerton_College/Math_100%3A_Liberal_Arts_Math_(Claassen_and_Ikeda)/10%3A_Appendix/10.06%3A_Problem_Solving/10.6.03%3A_Proportions_and_Rates\( \begin{array} {ll} {20\text{ seconds } \cdot \frac{1 \text { minute }}{60 \text { seconds }} \cdot \frac{1 \text { hour }}{60 \text { minutes }}=\frac{1}{180}\text{ hour}} & {\text{Now we can multi...\( \begin{array} {ll} {20\text{ seconds } \cdot \frac{1 \text { minute }}{60 \text { seconds }} \cdot \frac{1 \text { hour }}{60 \text { minutes }}=\frac{1}{180}\text{ hour}} & {\text{Now we can multiply by the }15\text{ miles/hr}} \\ {\frac{1}{180} \text { hour } \cdot \frac{15 \text { miles }}{\text { Ihour }}=\frac{1}{12} \text { mile }} & {\text{Now we can convert to feet}} \\ {\frac{1}{12} \text { mile } \cdot \frac{5280 \text { feet }}{1 \text { mile }}=440 \text { feet}} & { } \end{array…
- https://math.libretexts.org/Courses/Coastline_College/Math_C180%3A_Calculus_I_(Everett)/05%3A_Integration/5.03%3A_The_Fundamental_Theorem_of_Calculus/5.3E%3A_Exercises_for_Section_5.362) The force of gravitational attraction between the Sun and a planet is \(F(θ)=\dfrac{GmM}{r^2(θ)}\), where \(m\) is the mass of the planet, \(M\) is the mass of the Sun, \(G\) is a universal consta...62) The force of gravitational attraction between the Sun and a planet is \(F(θ)=\dfrac{GmM}{r^2(θ)}\), where \(m\) is the mass of the planet, \(M\) is the mass of the Sun, \(G\) is a universal constant, and \(r(θ)\) is the distance between the Sun and the planet when the planet is at an angle \(θ\) with the major axis of its orbit.
- https://math.libretexts.org/Courses/Coastline_College/Math_C160%3A_Introduction_to_Statistics_(Tran)/00%3A_Front_Matter/04%3A_LicensingA detailed breakdown of this resource's licensing can be found in Back Matter/Detailed Licensing.
- https://math.libretexts.org/Courses/Coastline_College/Math_C180%3A_Calculus_I_(Tran)/03%3A_Derivatives/3.02%3A_Defining_the_Derivative/3.2E%3A_Exercises_for_Section_3.1For exercises 1 - 10, use the equation \( m_{\text{sec}}=\frac{f(x)−f(a)}{x−a} \) to find the slope of the secant line between the values \(x_1\) and \(x_2\) for each function \(y=f(x)\). For the func...For exercises 1 - 10, use the equation \( m_{\text{sec}}=\frac{f(x)−f(a)}{x−a} \) to find the slope of the secant line between the values \(x_1\) and \(x_2\) for each function \(y=f(x)\). For the functions \(y=f(x)\) in exercises 21 - 30, find \(f′(a)\) using the equation \( \displaystyle f′(a)=\lim_{x→a}\frac{f(x)−f(a)}{x−a} \). For the following exercises, use the limit definition of derivative to show that the derivative does not exist at \(x=a\) for each of the given functions.
- https://math.libretexts.org/Bookshelves/Analysis/Mathematical_Analysis_(Zakon)/02%3A_Real_Numbers_and_Fields/2.05%3A_Some_Consequences_of_the_Completeness_AxiomThis proves our last assertion and shows that \(n o y \in F\) can be a right bound of \(N(\) for \(y<n \in N),\) or a left bound of \(J(\) for \(y>-m \in J). \square\) Now, by Theorem 2 of \(§§5-6, A+...This proves our last assertion and shows that \(n o y \in F\) can be a right bound of \(N(\) for \(y<n \in N),\) or a left bound of \(J(\) for \(y>-m \in J). \square\) Now, by Theorem 2 of \(§§5-6, A+m\) has a minimum; call it \(p .\) As \(p\) is the least of all sums \(x+m, p-m\) is the least of all \(x \in A ;\) so \(p-m=\min A\) exists, as claimed.
- https://math.libretexts.org/Bookshelves/Linear_Algebra/Matrix_Analysis_(Cox)/08%3A_The_Eigenvalue_Problem/8.05%3A_The_Eigenvalue_Problem-_Examples\[\begin{array}{ccc} {P_{1} = e_{1}(e_{1}^{T}e_{1})^{-1}e_{1}^{T}}&{and}&{P_{2} = e_{2}(e_{2}^{T}e_{2})^{-1}e_{2}^{T}} \end{array} \nonumber\] It is not the square root of the sum of squares of its co...\[\begin{array}{ccc} {P_{1} = e_{1}(e_{1}^{T}e_{1})^{-1}e_{1}^{T}}&{and}&{P_{2} = e_{2}(e_{2}^{T}e_{2})^{-1}e_{2}^{T}} \end{array} \nonumber\] It is not the square root of the sum of squares of its components but rather the square root of the sum of squares of the magnitudes of its components. \[\begin{array}{ccc} {P_{1} = e_{1}(e_{1}^{H}e_{1})^{-1}e_{1}^{H}}&{and}&{P_{2} = e_{2}(e_{2}^{H}e_{2})^{-1}e_{2}^{H}} \end{array} \nonumber\]
