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3 years ago

Parallelogram CDEF is formed inside of rectangle AEBC to create triangle ADC and triangle EBF as shown in the model below.

Mathematics

1 answer:

o-na [289]3 years ago

5 0

Answer:

Step-by-step explanation:

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What number has 62 hundreds and 5 tens

fomenos

Answer:

6250

Step-by-step explanation:

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3 years ago

Read 2 more answers

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:

Fundamental Theorem of Calculus

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

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:

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

Integration by substitution

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

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

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

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

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

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

Substitute 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:

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2 years ago

Box 1 measures 6cm in width, 12cm in length, and 4cm in height. The height is doubled to 8cm to produce a larger box, Box 2 Whic

arlik [135]

Step-by-step explanation:

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The height is doubled to 8cm to produce a larger box, Box 2 Which phrase accurately describes the volume of Box 2

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3 years ago

What is the following product? Sqr5x^8y^2 sqr10x^3 sqr12y

PSYCHO15rus [73]

Step-by-step explanation:

$\sqrt{5 {x}^{8} {y}^{2} } \times \sqrt{10 {x}^{3} } \times \sqrt{12y} \\ \\ = {x}^{4} y \sqrt{5} \times x \sqrt{10x} \times 2 \sqrt{3y} \\ \\ = 2 {x}^{5} y \sqrt{5 \times 10x \times 3y} \\ \\ = 2 {x}^{5} y \sqrt{5 \times 5 \times 2x \times 3y} \\ \\ = 2 {x}^{5} y \times 5\sqrt{ 2x \times 3y} \\ \\ = 10 {x}^{5} y \sqrt{6xy} \\$

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4 years ago

Find the sum of fourteen and eight tenths and 276.93

vitfil [10]

Answer:

299.03

Step-by-step explanation:

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