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91Ó°ÊÓ

$$h-0.1$$ $$h=0.05$$

Short Answer

Expert verified
The result is -0.05.

Step by step solution

01

Understand the Exercise

We are given the expression \( h - 0.1 \) and the value \( h = 0.05 \). The task is to substitute the given value of \( h \) into the expression and evaluate it.
02

Substitute the Value

Substitute \( h = 0.05 \) into the expression \( h - 0.1 \). This gives us: \( 0.05 - 0.1 \).
03

Perform the Subtraction

Calculate the subtraction: \( 0.05 - 0.1 \). This involves subtracting a larger number from a smaller one, resulting in a negative number. \( 0.05 - 0.1 = -0.05 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Substitution Method
The substitution method is very useful when solving algebraic expressions and equations. It primarily involves replacing a variable in an expression with its given numerical value. This helps simplify the expression, allowing you to perform further calculations. In our exercise, we begin with the expression \( h - 0.1 \), where we know that \( h = 0.05 \). By using substitution, we can directly replace \( h \) with its value. Thus, the expression \( h - 0.1 \) becomes \( 0.05 - 0.1 \). Using substitution is a common technique in algebra. It is crucial when values of variables are known or need to be replaced temporarily to simplify problems. Always ensure the correct value is used for the appropriate variable.
Negative Numbers
Negative numbers can be a source of difficulty, but with understanding, they become a powerful tool. They represent values less than zero and are often depicted with a minus sign (\(-\)). In the given exercise, when performing the operation \( 0.05 - 0.1 \), you are subtracting a larger number from a smaller one. This results in a negative number. Here, \( 0.05 - 0.1 = -0.05 \). When working with negative numbers, keep the following in mind:
  • Adding a negative number is the same as subtracting its absolute value.
  • Subtracting a negative number is the same as adding its absolute value.
  • Multiplying or dividing two negative numbers results in a positive number.
These rules help in understanding and efficiently manipulating negative numbers in algebraic operations.
Arithmetic Operations
Arithmetic operations form the backbone of mathematics, and they include addition, subtraction, multiplication, and division. In algebra, these operations are applied to numbers and algebraic expressions. In our exercise, we performed the subtraction of two numbers: \( 0.05 \) and \( 0.1 \). This operation is straightforward, yet it’s important to align decimal points carefully:- When subtracting \( 0.1 \) (equal to \( 0.10 \)) from \( 0.05 \), think of borrowing across the decimal, just like traditional subtraction. - Each number after the decimal point must have an equal number of digits when performing arithmetic operations to avoid mistakes.With practice, arithmetic operations become second nature. Remember, always double-check your calculations to ensure accuracy, especially when negatives or multiple decimal places are involved.

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Most popular questions from this chapter

(a) Solving \(v_{t}=\sqrt{m g / k}\) for \(k\) we obtain \(k=m g / v_{t}^{2} .\) The differential equation then becomes \\[ m \frac{d v}{d t}=m g-\frac{m g}{v_{t}^{2}} v^{2} \\] or \(\frac{d v}{d t}=g\left(1-\frac{1}{v_{t}^{2}} v^{2}\right)\) Separating variables and integrating gives \\[ v_{t} \tanh ^{-1} \frac{v}{v_{t}}=g t+c_{1} \\] The initial condition \(v(0)=0\) implies \(c_{1}=0,\) so \\[ v(t)=v_{t} \tanh \frac{g t}{v_{t}} \\] We find the distance by integrating: \\[ s(t)=\int v_{t} \tanh \frac{g t}{v_{t}} d t=\frac{v_{t}^{2}}{g} \ln \left(\cosh \frac{g t}{v_{t}}\right)+c_{2} \\] The initial condition \(s(0)=0\) implies \(c_{2}=0,\) so \\[ s(t)=\frac{v_{t}^{2}}{g} \ln \left(\cosh \frac{g t}{v_{t}}\right) \\] In 25 seconds she has fallen \(20,000-14,800=5,200\) feet. Using a CAS to solve \\[ 5200=\left(v_{t}^{2} / 32\right) \ln \left(\cosh \frac{32(25)}{v_{t}}\right) \\] for \(v_{t}\) gives \(v_{t} \approx 271.711 \mathrm{ft} / \mathrm{s} .\) Then \\[ s(t)=\frac{v_{t}^{2}}{g} \ln \left(\cosh \frac{g t}{v_{t}}\right)=2307.08 \ln (\cosh 0.117772 t) \\] (b) \(\operatorname{At} t=15, s(15)=2,542.94 \mathrm{ft}\) and \(v(15)=s^{\prime}(15)=256.287 \mathrm{ft} / \mathrm{sec}.\)

For \(y^{\prime}+2 y=f(x)\) an integrating factor is \(e^{2 x}\) so that $$y e^{2 x}=\left\\{\begin{array}{ll} \frac{1}{2} e^{2 x}+c_{1}, & 0 \leq x \leq 3 \\ c_{2}, & x>3 \end{array}\right.$$ If \(y(0)=0\) then \(c_{1}=-1 / 2\) and for continuity we must have \(c_{2}=\frac{1}{2} e^{6}-\frac{1}{2}\) so that $$y=\left\\{\begin{array}{ll} \frac{1}{2}\left(1-e^{-2 x}\right), & 0 \leq x \leq 3 \\ \frac{1}{2}\left(e^{6}-1\right) e^{-2 x}, & x>3 \end{array}\right.$$

Write the differential equation as \(d P / d t-a P=-b P^{2}\) and let \(u=P^{-1}\) or \(P=u^{-1}\). Then $$\frac{d p}{d t}=-u^{-2} \frac{d u}{d t},$$ and substituting into the differential equation, we have $$-u^{-2} \frac{d u}{d t}-a u^{-1}=-b u^{-2} \quad \text { or } \quad \frac{d u}{d t}+a u=b.$$ The latter differential equation is linear with integrating factor \(e^{\int a d t}=e^{a t},\) so $$\frac{d}{d t}\left[e^{a t} u\right]=b e^{a t}$$ and $$\begin{aligned} e^{a t} u &=\frac{b}{a} e^{a t}+c \\ e^{a t} P^{-1} &=\frac{b}{a} e^{a t}+c \\ P^{-1} &=\frac{b}{a}+c e^{-a t} \\ P &=\frac{1}{b / a+c e^{-a t}}=\frac{a}{b+c_{1} e^{-a t}}. \end{aligned}$$

From \(y^{\prime}-\left(1+\frac{1}{x}\right) y=y^{2}\) and \(w=y^{-1}\) we obtain \(\frac{d w}{d x}+\left(1+\frac{1}{x}\right) w=-1\). An integrating factor is \(x e^{x}\) so that \( x e^{x} w=-x e^{x}+e^{x}+c \text { or } y^{-1}=-1+\frac{1}{x}+\frac{c}{x} e^{-x}.\)

For \(y^{\prime}+2 x y=f(x)\) an integrating factor is \(e^{x^{2}}\) so that $$y e^{x^{2}}=\left\\{\begin{array}{ll} \frac{1}{2} e^{x^{2}}+c_{1}, & 0 \leq x \leq 1 \\ c_{2}, & x>1 \end{array}\right.$$ If \(y(0)=2\) then \(c_{1}=3 / 2\) and for continuity we must have \(c_{2}=\frac{1}{2} e+\frac{3}{2}\) so that $$y=\left\\{\begin{array}{ll} \frac{1}{2}+\frac{3}{2} e^{-x^{2}}, & 0 \leq x \leq 1 \\ \left(\frac{1}{2} e+\frac{3}{2}\right) e^{-x^{2}}, & x>1 \end{array}\right.$$

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