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2 years ago

208=(1+5x)Need help, i don't get it.

Mathematics

2 answers:

Semmy [17]2 years ago

8 0

Hello there!

First you had to switch sides of equation form.

$1+5x=208$

Then you subtract by one from both sides of equation form.

$1+5x-1=208-1$

Simplify.

$5x=207$

Next, you had to divide by five from both sides of equation form.

$\frac{5x}{5}= \frac{207}{5}$

Finally, simplify by equation form.

$x=\frac{207}{5}$

$207/5=41.4$

You can also round up to the nearest tenths is 41.0 it will stay.

Final answer: $\boxed{x=207/5}$ or $\boxed{x=41.4=41.0}$

Hope this helps!

And thank you for posting your question at here on brainly, and have a great day.

-Charlie

Snezhnost [94]2 years ago

3 0

X equals 41.4 I think

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3 years ago

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$\bf y=\cfrac{2x^2-10x}{\sqrt{x}}\implies y=\cfrac{2x^2-10x}{x^{\frac{1}{2}}} \\\\\\ \cfrac{dy}{dx}=\stackrel{\textit{quotient rule}}{\cfrac{(4x-10)(\sqrt{x})~~-~~(2x^2-10x)\left( \frac{1}{2}x^{-\frac{1}{2}} \right)}{\left( x^{\frac{1}{2}} \right)^2}} \\\\\\ \cfrac{dy}{dx}=\cfrac{(4x-10)(\sqrt{x})~~-~~(2x^2-10x)\left( \frac{1}{2\sqrt{x}} \right)}{\left( x^{\frac{1}{2}} \right)^2} \\\\\\ \cfrac{dy}{dx}=\cfrac{(4x-10)(\sqrt{x})~~-~~\left( \frac{2x^2-10x}{2\sqrt{x}} \right)}{x}$

$\bf\cfrac{dy}{dx}=\cfrac{(4x-10)(\sqrt{x})~~-~~\left( \frac{2x^2-10x}{2\sqrt{x}} \right)}{x} \\\\\\ \cfrac{dy}{dx}=\cfrac{ \frac{(4x-10)(\sqrt{x})(2\sqrt{x})~~-~~(2x^2-10x)}{2\sqrt{x}}}{x} \\\\\\ \cfrac{dy}{dx}=\cfrac{(4x-10)(\sqrt{x})(2\sqrt{x})~~-~~(2x^2-10x)}{2x\sqrt{x}}$

$\bf \cfrac{dy}{dx}=\cfrac{(4x-10)2x~~-~~(2x^2-10x)}{2x\sqrt{x}}\implies \cfrac{dy}{dx}=\cfrac{8x^2-20x~~-~~(2x^2-10x)}{2x\sqrt{x}} \\\\\\ \cfrac{dy}{dx}=\cfrac{8x^2-20x~~-~~2x^2+10x}{2x\sqrt{x}} \implies \cfrac{dy}{dx}=\cfrac{6x^2-10x}{2x\sqrt{x}} \\\\\\ \cfrac{dy}{dx}=\cfrac{2x(3x-5)}{2x\sqrt{x}}\implies \cfrac{dy}{dx}=\cfrac{3x-5}{\sqrt{x}}$

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