3 years ago

Work out the acceleration, in km/h^2 in the first 15mins

Mathematics

1 answer:

kodGreya [7K]3 years ago

7 0

B is the correct answer if it wrong am sorry

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Find the derivative. <br> y = 5/ (2root4(x^3))

DedPeter [7]

$\bf y=\cfrac{5}{2\sqrt[4]{x^3}}\implies y=\cfrac{5}{2}\cdot x^{-\frac{3}{4}}\\\\
-----------------------------\\\\
\cfrac{dy}{dx}=\cfrac{5}{2}\left( -\cfrac{3}{4}x^{-\frac{3}{4}-1} \right)\implies 
\cfrac{dy}{dx}=\cfrac{5}{2}\left( -\cfrac{3}{4}x^{-\frac{7}{4}} \right)
\\\\\\
 
\cfrac{dy}{dx}=-\cfrac{15}{8}\cdot x^{-\frac{7}{4}} 
\implies 
\cfrac{dy}{dx}=-\cfrac{15}{8}\cdot \cfrac{1}{x^{\frac{7}{4}} }
\\\\\\
$

$\bf
\cfrac{dy}{dx}=-\cfrac{15}{8x^{\frac{7}{4}}}\implies 
\cfrac{dy}{dx}=-\cfrac{15}{8\sqrt[4]{x^7}}\\\\\\ \cfrac{dy}{dx}=-\cfrac{15}{8x\sqrt[4]{x^3}}
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3 0

3 years ago

If the initial figure is 10 and then it is dilated with a scale factor of 2 what's the perimeter of dilated figure?

Gelneren [198K]

Answer:

Step-by-step explanation:

Initial figure times scale factor = result

10 times 2 = 20

I hope this helps!

8 0

2 years ago

27-4(26-2×3)to the 2nd power