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Iteru [2.4K]
3 years ago
14

Fully simplify −10x3y2(9x)

Mathematics
1 answer:
lbvjy [14]3 years ago
5 0

Answer:

Step-by-step explanation:

-10*9*x^3*x*y^2 = -90x^4y^2

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Passes through (4, 2) and (4,-5) in standard form
elena-s [515]

Answer:

x+4=0

Step-by-step explanation:

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3 years ago
What is the perimeter of the rectangle? Plato
charle [14.2K]

it depends how long the width and the length is. If the length is 5 cm and the width is 4 cm the perimeter will be 18cm. All you have to do is 5 doubled because there 2 sides for the length and 4 doubled because there are 2 sides for the width. I hope I have answered your question. XD

3 0
3 years ago
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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

4 0
2 years ago
Someone help me solve this and could you explain why too?
yulyashka [42]

Answer:

3.5

Step-by-step explanation:

Well its easiest to just plug in the answers and see like 2 times 3.5 is 7.

Another way is saying 7 divided by 2 is 3.5

then u check to make sure its the same for all the numbers.

Which it is because 21 divided by 6 is also 3.5 :)

5 0
3 years ago
What is the sum of 5.3 x 10^5 and 3.8 x 10^4 in scientific notation?
Sedaia [141]
5.3×10^5=53×10^4
53×10^4+3.8×10^4
=56.8×10^4
=5.68×10^5. Hope it help!
3 0
3 years ago
Read 2 more answers
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