3 years ago

Solve. 2x + 4 3 B) x < 3 B) x 1/3D) x < 1/3

Mathematics

1 answer:

kakasveta [241]3 years ago

7 0

Answer:

Step-by-step explanation:

You might be interested in

What is the simplified form of each expression? (– h 4)5 A. – h 9 B. h 1 C. h 20 D. – h 20

Katena32 [7]

8 0

3 years ago

7-18 use part 1 of the fundamental theorem of calculus to find the derivative of the function.

blondinia [14]

$\displaystyle h(x)=\int\limits_{1}^{\sqrt{x}}~\cfrac{z^2}{z^4+1}dz~\hspace{10em}\cfrac{dh}{du}\cdot \stackrel{chain~rule}{\cfrac{du}{dx}\implies \cfrac{dh}{dx}} \\\\[-0.35em] ~\dotfill\\\\ u=\sqrt{x}\implies \cfrac{du}{dx}=\cfrac{1}{2\sqrt{x}} \\\\[-0.35em] ~\dotfill$

$\cfrac{dh}{dx}\implies \displaystyle \cfrac{d}{du}\left[ \int\limits_{1}^{u}~\cfrac{z^2}{z^4+1}dz \right]\cdot \cfrac{1}{2\sqrt{x}}\implies \left[ \cfrac{u^2}{u^4+1} \right]\cdot \cfrac{1}{2\sqrt{x}} \\\\\\ \stackrel{substituting~back}{\left[ \cfrac{(\sqrt{x})^2}{(\sqrt{x})^4+1} \right]\cdot \cfrac{1}{2\sqrt{x}}}\implies \cfrac{x}{x^2+1}\cdot \cfrac{1}{2\sqrt{x}}\implies \cfrac{\sqrt{x}}{2x^2+2}$