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luda_lava [24]
3 years ago
8

What is (f + g)(x)? f(x)=x^3−x g(x)=x^3+2x^2−10

Mathematics
1 answer:
Leya [2.2K]3 years ago
4 0
(f+g)(x)=f(x)+g(x) = x³-x +x³ +2x²-10=2x³ + 2x²-x-10
(f+g)(x)=2x³ + 2x²-x-10
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3. [6x - 2y=2.<br> (2 + 6x=y
Hunter-Best [27]

Answer:

3. [6x - 2y=2.

(2 + 6x=y

Step-by-step explanation:

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Write 28 + 24 as a product of two factors using the GCF and the distributive property
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4 x 13

Step-by-step explanation:

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A string of decorative lights is 25 feet long. The first light on the string is 16 inches from the plug. The lights on the strin
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3 years ago
Evaluate the interval (Calculus 2)
Darya [45]

Answer:

2 \tan (6x)+2 \sec (6x)+\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{12}{1-\sin (6x)}\:\:\text{d}x

\boxed{\begin{minipage}{5 cm}\underline{Terms multiplied by constants}\\\\$\displaystyle \int a\:\text{f}(x)\:\text{d}x=a \int \text{f}(x) \:\text{d}x$\end{minipage}}

If the terms are multiplied by constants, take them outside the integral:

\implies 12\displaystyle \int \dfrac{1}{1-\sin (6x)}\:\:\text{d}x

Multiply by the conjugate of 1 - sin(6x) :

\implies 12\displaystyle \int \dfrac{1}{1-\sin (6x)} \cdot \dfrac{1+\sin(6x)}{1+\sin(6x)}\:\:\text{d}x

\implies 12\displaystyle \int \dfrac{1+\sin(6x)}{1-\sin^2(6x)} \:\:\text{d}x

\textsf{Use the identity} \quad \sin^2 x+ \cos^2 x=1:

\implies \sin^2 (6x) + \cos^2 (6x)=1

\implies \cos^2 (6x)=1- \sin^2 (6x)

\implies 12\displaystyle \int \dfrac{1+\sin(6x)}{\cos^2(6x)} \:\:\text{d}x

Expand:

\implies 12\displaystyle \int \dfrac{1}{\cos^2(6x)}+\dfrac{\sin(6x)}{\cos^2(6x)} \:\:\text{d}x

\textsf{Use the identities }\:\: \sec \theta=\dfrac{1}{\cos \theta} \textsf{ and } \tan\theta=\dfrac{\sin \theta}{\cos \theta}:

\implies 12\displaystyle \int \sec^2(6x)+\dfrac{\tan(6x)}{\cos(6x)} \:\:\text{d}x

\implies 12\displaystyle \int \sec^2(6x)+\tan(6x)\sec(6x) \:\:\text{d}x

\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}}

\boxed{\begin{minipage}{6 cm}\underline{Integrating $ \sec kx \tan kx$}\\\\$\displaystyle \int  \sec kx \tan kx\:\text{d}x= \dfrac{1}{k}\sec kx\:\:(+\text{C})$\end{minipage}}

\implies 12 \left[\dfrac{1}{6} \tan (6x)+\dfrac{1}{6} \sec (6x) \right]+\text{C}

Simplify:

\implies \dfrac{12}{6} \tan (6x)+\dfrac{12}{6} \sec (6x)+\text{C}

\implies 2 \tan (6x)+2 \sec (6x)+\text{C}

Learn more about indefinite integration here:

brainly.com/question/27805589

brainly.com/question/28155016

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