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Bad White [126]
3 years ago
9

Which of these object does not represent a plane

Mathematics
1 answer:
Hatshy [7]3 years ago
8 0

Answer:

D. star in the sky

Step-by-step explanation:

To be honest, none of them represent the plane but I think that's a better answer. OwO

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Suppose you are asked to find the area of a rectangle that is 2.1-cm wide by 5.6-cm long. Your calculator answer would be 11.76
yKpoI14uk [10]

Answer:

12 cm²

Step-by-step explanation:

Length of rectangle = 5.6 cm

Width of rectangle = 2.1 cm

Area of rectangle = Length of rectangle×Width of rectangle

⇒Area of rectangle = 5.6×2.1

⇒Area of rectangle = 11.76 cm²

11.76 has 4 significant figures in order to write this term in 2 significant terms we round of the term

The last digit in the decimal place is 6. Now, 6≥5 so we round the next digit to 8 we get

11.8

Now the last digit in the decimal place is 8. Now, 8≥5 so we round the next digit to 2 we get

12

∴ Hence the area of the rectangle when rounded to 2 significant figures is 12 cm²

5 0
3 years ago
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kirza4 [7]
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6 0
3 years ago
What’s the answer???
andre [41]

Answer:

567 i think

Step-by-step explanation:

3 0
2 years ago
Read 2 more answers
Find dy/dx if y =x^3+5x+2/x²-1
stiks02 [169]

<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 –

  • f(x) = x^3+5x+2
  • g(x) = x^2-1

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 }}}}}\\\\

7 0
2 years ago
40.4 Kilograms round to the nearest hundreth​
Romashka-Z-Leto [24]

Answer: 40.4 rounded to the nearest hundredth would be 40.40

Step-by-step explanation: There is no oneth in decimals so it goes from tenth, to hundredth, to thousandth. Since there is no number in the hundredth or thousandth's space, you would put a zero in both spots. zero in the thousandths is NOT greater than 5 so the zero in the hundredths space will stay as a zero, making the answer 40.40

3 0
3 years ago
Read 2 more answers
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