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Dominik [7]
4 years ago
14

Match the following items.

Mathematics
2 answers:
Gwar [14]4 years ago
8 0

Answer:

1. Prefix meaning 1/1,000 → milli

2. Prefix measuring 1,000 → kilo

3. Prefix meaning 1/100 → centi

4. Base unit of distance → meter

5. Base unit of mass  → kilogram

6. Base unit of volume → Liter

Step-by-step explanation:

1. Prefix meaning 1/1,000 → milli

Milli is a unit prefix in a metric system which denotes the factor of a10^{-3}

2. Prefix measuring 1,000 → kilo

Kilo is a unit prefix in a metric system which denotes the factor of a 10^3.

3. Prefix meaning 1/100 → centi

Centi is a unit prefix in a metric system which denotes the factor of a 10^{-2}

Base unit are those units which are defined independently and not by combination of other units.

4. Base unit of distance → meter

Distance is measured in meter

5. Base unit of mass  → kilogram

Mass is measured in kilogram

6. Base unit of volume → Liter

Volume is measured in liters

ASHA 777 [7]4 years ago
7 0
1 is milli. 2 is kilo. 3 is centi. 4 is meter. 5 is kilogram. 6 is liter.
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Find dy/dx if y =x^3+5x+2/x²-1
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<u>Differentiate using the Quotient Rule</u> –

\qquad\pink{\twoheadrightarrow \sf \dfrac{d}{dx} \bigg[\dfrac{f(x)}{g(x)} \bigg]= \dfrac{ g(x)\:\dfrac{d}{dx}\bigg[f(x)\bigg] -f(x)\dfrac{d}{dx}\:\bigg[g(x)\bigg]}{g(x)^2}}\\

According to the given question, we have –

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Let's solve it!

\qquad\green{\twoheadrightarrow \bf \dfrac{d}{dx}\bigg[ \dfrac{x^3+5x+2 }{x^2-1}\bigg]} \\

\qquad\twoheadrightarrow \sf \dfrac{(x^2-1) \dfrac{d}{dx}(x^3+5x+2) - ( x^3+5x+2)  \dfrac{d}{dx}(x^2-1)}{(x^2-1)^2 }\\

\qquad\twoheadrightarrow \sf \dfrac{(x^2-1)(3x^2+5)  -  ( x^3+5x+2) 2x}{(x^2-1)^2 }\\

\qquad\pink{\sf \because \dfrac{d}{dx} x^n = nx^{n-1} }\\

\qquad\twoheadrightarrow \sf \dfrac{3x^4+5x^2-3x^2-5-(2x^4+10x^2+4x)}{(x^2-1)^2 }\\

\qquad\twoheadrightarrow \sf \dfrac{3x^4+5x^2-3x^2-5-2x^4-10x^2-4x}{(x^2-1)^2 }\\

\qquad\green{\twoheadrightarrow \bf \dfrac{x^4-8x^2-4x-5}{(x^2-1)^2 }}\\

\qquad\pink{\therefore  \bf{\green{\underline{\underline{\dfrac{d}{dx} \dfrac{x^3+5x+2 }{x^2-1}}  =  \dfrac{x^4-8x^2-4x-5}{(x^2-1)^2 }}}}}\\\\

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