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3 years ago

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Mike spent half of his allowance going to the movies.He washed the family car and earned 9 dollars.What is his weekly allowance

is he ended with 18 dollars

1 answer:

Bezzdna [24]3 years ago

7 0

Answer:

18

Step-by-step explanation:

If he spends half his money then washes the family car and he ends up with 18 dollars that means he spent 9 witch is half of 18 and he adds 9 more dollars which is why his allowance is 18 dollars

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Kol and Kara solve this problem in two different ways the perimeter of a rectangle is 46 centimeters it’s length is 8 centimeter

JulsSmile [24]

Answer:

15cm

Step-by-step explanation:

because length is given as 8 cm and perimeter is length+width+length+width

so if you take 8cmX2 since its taken twice to get the perimeter and 15cmX2 you get 46cm which originally the question said was the peemiter

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3 years ago

Evaluate the integral, show all steps please!

Aloiza [94]

Answer:

$\displaystyle \int \dfrac{1}{(9-x^2)^{\frac{3}{2}}}\:\:\text{d}x=\dfrac{x}{9\sqrt{9-x^2}} +\text{C}$

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:

$\displaystyle \int \dfrac{1}{(9-x^2)^{\frac{3}{2}}}\:\:\text{d}x$

Rewrite 9 as 3² and rewrite the 3/2 exponent as square root to the power of 3:

$\implies \displaystyle \int \dfrac{1}{\left(\sqrt{3^2-x^2}\right)^3}\:\:\text{d}x$

<u>Integration by substitution</u>

<u />

<u /> $\boxed{\textsf{For }\sqrt{a^2-x^2} \textsf{ use the substitution }x=a \sin \theta}$

$\textsf{Let }x=3 \sin \theta$

$\begin{aligned}\implies \sqrt{3^2-x^2} & =\sqrt{3^2-(3 \sin \theta)^2}\\ & = \sqrt{9-9 \sin^2 \theta}\\ & = \sqrt{9(1-\sin^2 \theta)}\\ & = \sqrt{9 \cos^2 \theta}\\ & = 3 \cos \theta\end{aligned}$

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

$\implies \dfrac{\text{d}x}{\text{d}\theta}=3 \cos \theta$

$\implies \text{d}x=3 \cos \theta\:\:\text{d}\theta$

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

$\begin{aligned}\displaystyle \int \dfrac{1}{(9-x^2)^{\frac{3}{2}}}\:\:\text{d}x & = \int \dfrac{1}{\left(\sqrt{3^2-x^2}\right)^3}\:\:\text{d}x\\\\& = \int \dfrac{1}{\left(3 \cos \theta\right)^3}\:\:3 \cos \theta\:\:\text{d}\theta \\\\ & = \int \dfrac{1}{\left(3 \cos \theta\right)^2}\:\:\text{d}\theta \\\\ & = \int \dfrac{1}{9 \cos^2 \theta} \:\: \text{d}\theta\end{aligned}$

Take out the constant:

$\implies \displaystyle \dfrac{1}{9} \int \dfrac{1}{\cos^2 \theta}\:\:\text{d}\theta$

$\textsf{Use the trigonometric identity}: \quad\sec^2 \theta=\dfrac{1}{\cos^2 \theta}$

$\implies \displaystyle \dfrac{1}{9} \int \sec^2 \theta\:\:\text{d}\theta$

$\boxed{\begin{minipage}{5 cm}\underline{Integrating $\sec^2 kx$}\\\\$\displaystyle \int \sec^2 kx\:\text{d}x=\dfrac{1}{k} \tan kx\:\:(+\text{C})$\end{minipage}}$

$\implies \displaystyle \dfrac{1}{9} \int \sec^2 \theta\:\:\text{d}\theta = \dfrac{1}{9} \tan \theta+\text{C}$

$\textsf{Use the trigonometric identity}: \quad \tan \theta=\dfrac{\sin \theta}{\cos \theta}$

$\implies \dfrac{\sin \theta}{9 \cos \theta} +\text{C}$

$\textsf{Substitute back in } \sin \theta=\dfrac{x}{3}:$

$\implies \dfrac{x}{9(3 \cos \theta)} +\text{C}$

$\textsf{Substitute back in }3 \cos \theta=\sqrt{9-x^2}:$

$\implies \dfrac{x}{9\sqrt{9-x^2}} +\text{C}$

Learn more about integration by substitution here:

brainly.com/question/28156101

brainly.com/question/28155016

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