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2 years ago

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If the coordinates for C' are (4.5, 4.5), which rule best represents the dilation that was applied to rectangle ABCD to create r

ectangle A'B'C'D'?

1 answer:

Dmitry_Shevchenko [17]2 years ago

7 0

The dilation of rectangle ABCD to create rectangle A'B'C'D' would change the size of the rectangle

The dilation rule is: (x,y) -> k(x,y)

<h3>How to determine the rule of dilation?</h3>

The coordinates of C' is given as:

C' = (4.5, 4.5)

The coordinate of point C is not given.

So, I will apply a general rule

The general rule of dilation is:

(x,y) -> k(x,y)

Where k represents the scale factor.

Assume that k = 3, the the rule would be
(x,y) -> 3(x,y)

Read more about dilation at:

brainly.com/question/3457976

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notsponge [240]

Using a linear function, it is found that the algebraic expression that shows the amount Ms. Jackson will pay for the small truck for the week is:

$y(m) = 0.6m + 340$

<h3>What is a linear function?</h3>

A linear function is modeled by:

$y = ax + b$

In which:

a is the slope, which is the rate of change, that is, by how much y changes when x changes by 1.

b is the y-intercept, which is the value of y when x = 0.

In this problem:

He pays an amount per mile, plus a flat weekly fee.

Researching the problem on the internet, it is found that the weekly fee for the small truck is of $340, hence $b = 340$ .

For each mile, he pays $0.6, hence the slope is of $a = 0.6$ .

Then, the expression is:

$y(m) = 0.6m + 340$

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3 years ago

How many per laps is that

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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})$\end{minipage}}$

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

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

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

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

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

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

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

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2 years ago

Solve the following decimals

Vikentia [17]

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2 years ago

How is 9 to the 0 power simplified

TiliK225 [7]

Answer:

1

Step-by-step explanation:

Anything to the power of 0 is 1

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3 years ago

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