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3 years ago

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Which algebraic expression is not equivalent to 13x + 10? A. 10(x + 1) + 3x B 4x + 3 + 8x + 7 0. C. 13 + 2x + 3x – 3 + 8x D. 20x

– 2 – 7x + 12

1 answer:

Temka [501]3 years ago

6 0

A, C, and D are all equivalent therefore the correct answer is.... (B). It is the only one that equals something else.

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Over a period of five days, the average time of a commuter train trip from one town to the city was 77 minutes. One trip was 71

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Average time of a commuter train trip from one town to the city = 77 minutes
Time taken by 1 trip = 71 minutes
Time taken by another trip = 74 minutes
Time taken by 3rd and 4th trip each = 78 minutes
Let us assume the time taken during the 5th trip = x
Then
(71 + 74 + 78 +78 +x)/5 = 77
301 + x = 77 * 5
301 + x = 385
x = 385 - 301
= 84 minutes
So the length of the fifth trip was 84 minutes. I hope the procedure is clear enough for you to understand.

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3 years ago

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Evaluate the following integral (Calculus 2) Please show step by step explanation!

Nuetrik [128]

Answer:

$4\ln \left| \dfrac{1}{3}\sqrt{9+(\ln x)^2} + \dfrac{1}{3}\ln x \right|+\text{C}$

Step-by-step explanation:

<u>Fundamental Theorem of Calculus</u>

$\displaystyle \int \text{f}(x)\:\text{d}x=\text{F}(x)+\text{C} \iff \text{f}(x)=\dfrac{\text{d}}{\text{d}x}(\text{F}(x))$

If differentiating takes you from one function to another, then integrating the second function will take you back to the first with a constant of integration.

Given indefinite integral:

$\displaystyle \int \dfrac{4}{x\sqrt{9+(\ln(x))^2}}\:\:\text{d}x$

Rewrite 9 as 3²:

$\implies \displaystyle \int \dfrac{4}{x\sqrt{3^2+(\ln(x))^2}}\:\:\text{d}x$

<u>Integration by substitution</u>

$\boxed{\textsf{For }\sqrt{a^2+x^2} \textsf{ use the substitution }x=a \tan\theta}$

$\textsf{Let } \ln x=3 \tan \theta$

$\begin{aligned}\implies \sqrt{3^2+(\ln x)^2} & =\sqrt{3^2+(3 \tan\theta)^2}\\ & = \sqrt{9+9\tan^2 \theta}\\ & = \sqrt{9(1+\tan^2 \theta)}\\ & = \sqrt{9\sec^2 \theta}\\ & = 3 \sec\theta\end{aligned}$

Find the derivative of ln x and rewrite it so that dx is on its own:

$\implies \ln x=3 \tan \theta$

$\implies \dfrac{1}{x}\dfrac{\text{d}x}{\text{d}\theta}=3 \sec^2\theta$

$\implies \text{d}x=3x \sec^2\theta\:\:\text{d}\theta$

<u>Substitute</u> everything into the original integral:

$\begin{aligned} \implies \displaystyle \int \dfrac{4}{x\sqrt{9+(\ln(x))^2}}\:\:\text{d}x & = \int \dfrac{4}{3x \sec \theta} \cdot 3x \sec^2\theta\:\:\text{d}\theta\\\\ & = \int 4 \sec \theta \:\: \text{d}\theta\end{aligned}$

Take out the constant:

$\implies \displaystyle 4 \int \sec \theta\:\:\text{d}\theta$

$\boxed{\begin{minipage}{7 cm}\underline{Integrating $\sec kx$}\\\\$\displaystyle \int \sec kx\:\text{d}x=\dfrac{1}{k} \ln \left| \sec kx + \tan kx \right|\:\:(+\text{C})$\end{minipage}}$

$\implies 4\ln \left| \sec \theta + \tan \theta \right|+\text{C}$

$\textsf{Substitute back in } \tan\theta=\dfrac{1}{3}\ln x :$

$\implies 4\ln \left| \sec \theta + \dfrac{1}{3}\ln x \right|+\text{C}$

$\textsf{Substitute back in } \sec\theta=\dfrac{1}{3}\sqrt{9+(\ln x)^2}:$

$\implies 4\ln \left| \dfrac{1}{3}\sqrt{9+(\ln x)^2} + \dfrac{1}{3}\ln x \right|+\text{C}$

Learn more about integration by trigonometric substitution here:

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2 years ago

An object is dropped from a height of 1700 ft above the ground. The function h=-16t^2+1700 gives the object’s height h in feet d

Softa [21]

The answer is
<span>a) 1000=-16t^2+1700, implies t² = -700 /-16, and t= 6.61s
b) </span><span>970= -16t^2+1700, </span><span>implies t² = -730 /-16, and t=6.75s

c)
reasonable domain of h
h is polynomial function, so its domain is R, (all real number)
its range
the inverse of h is h^-1 = sqrt (1700- t / 16), and its domain is </span>
<span><span><span>1700- t / 16>=0, so t <1700,
the range of h is I= ]-infinity, 1700]</span> </span> </span>

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Answer:

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