/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 81 \(\mathrm{s}=30\left(1-\mathrm{e... [FREE SOLUTION] | 91影视

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Chapter 5: Problem 81

\(\mathrm{s}=30\left(1-\mathrm{e}^{\mathrm{ln}}\right)\) \(5=30\left(\mathrm{l}-\mathrm{e}^{k}\right)\) \(\mathrm{e}^{k}=\frac{5}{6}\) \(\mathrm{k}=\ln (5 / 6)\)

Short Answer

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#tag_title# Step 7: Summarize the solution #tag_content# We have solved the given equations and found the variables as follows: \(s = 30(1 - A)\) \(\mathrm{l} = 1\) \(k = \ln \left(\frac{5}{6}\right)\) These are the values of the variables for the given system of equations.

Step by step solution

01

Rewrite the first equation

Rewrite the equation \(s = 30\left(1-\mathrm{e}^{\mathrm{ln}}\right)\) as \(s = 30(1 - e^{\ln{A}})\).
02

Simplify using logarithmic properties

Recall the property of logarithms: \(e^{\ln{A}} = A\). Using this property, we can simplify the first equation: \(s = 30(1 - A)\)
03

Address the other system of equations

Now, let's focus on the system of equations: \(5 = 30\left(\mathrm{l}-\mathrm{e}^{k}\right)\) \(\mathrm{e}^{k} = \frac{5}{6}\) \(\mathrm{k} = \ln (5 / 6)\)
04

Solve for \(k\)

The third equation shows that \(e^k = \frac{5}{6}\). The fourth equation has \(k\) expressed as a logarithm. So the fourth equation gives us: \(k = \ln \left(\frac{5}{6}\right)\)
05

Connect \(k\) and the first system of equations

Since we have found \(k\) in terms of a logarithm, we can substitute the value of \(k\) into the second equation: \(5 = 30\left(\mathrm{l}-\mathrm{e}^{\ln \left(\frac{5}{6}\right)}\right)\) Then, using the property of logarithms we used in step 2, we get: \(5 = 30\left(\mathrm{l} - \frac{5}{6}\right)\)
06

Solve for \(\mathrm{l}\)

Now, we just need to solve for the variable \(\mathrm{l}\): \(5 = 30\left(\mathrm{l}-\frac{5}{6}\right)\) \(\frac{5}{30} = \mathrm{l}-\frac{5}{6}\) \(\frac{1}{6} = \mathrm{l}-\frac{5}{6}\) \(\mathrm{l} = \frac{1}{6}+\frac{5}{6} = 1\) So, we have found the variable \(\mathrm{l} = 1\).

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Most popular questions from this chapter

Slope of secant \(=\frac{9 a^{2}-a^{2}}{3 a-a}=4 a\) \(\frac{d y}{d x}=2 x=4 a\) \(x=2 a\) \(y=4 a^{2}\)

\(y=\left(\frac{x}{2}-a\right)^{2}+a-2\) \(4(y-(a-2))=(x-2 a)^{2}\) Vertex \(\Rightarrow \mathrm{h}=2 \mathrm{a}, \mathrm{k}=\mathrm{a}-2\) Locus of vertex \(\Rightarrow \mathrm{y}=\frac{\mathrm{x}}{2}-2\) \(\Rightarrow 2 y=x-4\)

A) Area of \(\mathrm{ABCD}=2 \mathrm{Ar}(\triangle \mathrm{ABC})\) \(=2 \times \frac{1}{2} \times \mathrm{BC} \times \mathrm{AB}\) \(=3 \times 2=6\) B) \(\mathrm{f}(\mathrm{x})=\frac{1}{\ln |\mathrm{x}|}\) is discontinous at \(x=0, \pm 1\) C) $f^{\prime}(x)=\lim _{n \rightarrow 0} \frac{f(x+n)-f(x)}{n}=\lim _{a \rightarrow 0} f(x)\left(\frac{f(n)-1)}{n}\right)$ \(=f(x) f^{\prime}(0)\) Put \(x=5\) \(f^{\prime}(5)=f(5) f^{\prime}(0)=6\) D) \(\tan \frac{3 \pi}{4}=-\frac{1}{f^{\prime}(3)}\) \(\Rightarrow f^{\prime}(3)=1\) \(A \rightarrow(R), B \rightarrow(Q), C \rightarrow(R), D \rightarrow(P)\)

eqn of normal \(y=m x-2 m-m^{3}\) It passes through \((6,0)\) \(\mathrm{m}^{3}-4 \mathrm{~m}=0\) \(\mathrm{m}\left(\mathrm{m}^{2}-4\right)=0\) \(\mathrm{m}=0, \pm 2\) Pts are \(\left(a m^{2}-2 a m\right)\) \((0,0),(4,-4),(4,4)\) Minimum distance \(=\sqrt{21}-\sqrt{5}\)

slope of normal \(\Rightarrow 3 x-y+3=0\) $$ x=0 \& y=3 $$ Pt of normal \(=(0,3)\) \(\frac{d y}{d x}=\frac{-1}{3}=f^{\prime}(0)\) $\lim _{x \rightarrow 0} \frac{x^{2}}{f\left(x^{2}\right)+4 f\left(7 x^{2}\right)-5 f\left(4 x^{2}\right)}$ $\lim _{x \rightarrow 0} \frac{2 x}{x\left[2 f^{\prime}\left(x^{2}\right)+56 f^{\prime}\left(7 x^{2}\right)-40 f^{\prime}\left(4 x^{2}\right)\right]}$ \(=\frac{2}{-6}=\frac{-1}{3}\)
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