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zlopas [31]
4 years ago
14

The approximate value of 5.5 million x 1.9 thousand is(scientific notation)

Mathematics
1 answer:
Andreas93 [3]4 years ago
5 0
The answer is:  " <span>1.045 * 10¹⁰  " .</span>
_____________________________________
Explanation:
______________________________________
 "5.5 million = 5,500,000 = 5.5 * 10 ⁶ "  ;
______________________________________
 "1.9 thousand = 1,900 = 1.9 * 10³  "   ;
______________________________________
   So;  "<span>5.5 million * 1.9 thousand" ; 
_________________________________________
  =  (</span>5.5 * 10 ⁶) * (1.9 * 10³) =  5.5 * 10⁶ * 1.9 * 10³ ;
 
     =  5.5 * 1.9 * 10⁶ * 10³ ;
 ___________________________________________
    Note the following properties of exponents:
____________________________________________
    xᵃ * xᵇ  =  x⁽ᵃ⁺ᵇ⁾ ;
____________________________________________

As such, " 10⁶ * 10³  = 10⁽ ⁶ ⁺ ³⁾  = 10⁹ " ; 
_____________________________________
So; rewrite:  
______________________________________
5.5 * 1.9 * 10⁶ * 10³  ;
_______________________________________
= 5.5 * 1.9 * 10⁹   ;
__________________________________
    5.5 * 1.9 = 10.45 ;
__________________________________
Rewrite as:  " 10.45 * 10⁹ " ;

In scientific notation, we need to rewrite as an single digit integer, followed by a decimal and other values (if applicable); following by * 10 raised to the appropriately adjusted exponential power.
_________________________________________________
As such, we write the value as:
________________________________________________
   " 1.045 * 10¹⁰  " . 
________________________________________________
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Evaluate the integral ∫2032x2+4dx. Your answer should be in the form kπ, where k is an integer. What is the value of k? (Hint: d
faltersainse [42]

Here is the correct computation of the question;

Evaluate the integral :

\int\limits^2_0 \ \dfrac{32}{x^2 +4}  \ dx

Your answer should be in the form kπ, where k is an integer. What is the value of k?

(Hint:  \dfrac{d \ arc \ tan (x)}{dx} =\dfrac{1}{x^2 + 1})

k = 4

(b) Now, lets evaluate the same integral using power series.

f(x) = \dfrac{32}{x^2 +4}

Then, integrate it from 0 to 2, and call it S. S should be an infinite series

What are the first few terms of S?

Answer:

(a) The value of k = 4

(b)

a_0 = 16\\ \\ a_1 = -4 \\ \\ a_2 = \dfrac{12}{5} \\ \\a_3 = - \dfrac{12}{7} \\ \\ a_4 = \dfrac{12}{9}

Step-by-step explanation:

(a)

\int\limits^2_0 \dfrac{32}{x^2 + 4} \ dx

= 32 \int\limits^2_0 \dfrac{1}{x+4}\  dx

=32 (\dfrac{1}{2} \ arctan (\dfrac{x}{2}))^2__0

= 32 ( \dfrac{1}{2} arctan (\dfrac{2}{2})- \dfrac{1}{2} arctan (\dfrac{0}{2}))

= 32 ( \dfrac{1}{2}arctan (1) - \dfrac{1}{2} arctan (0))

= 32 ( \dfrac{1}{2}(\dfrac{\pi}{4})- \dfrac{1}{2}(0))

= 32 (\dfrac{\pi}{8}-0)

= 32 ( (\dfrac{\pi}{8}))

= 4 \pi

The value of k = 4

(b) \dfrac{32}{x^2+4}= 8 - \dfrac{3x^2}{2^1}+ \dfrac{3x^4}{2^3}- \dfrac{3x^6}{2x^5}+ \dfrac{3x^8}{2^7} -...  \ \ \ \ \ (Taylor\ \ Series)

\int\limits^2_0  \dfrac{32}{x^2+4}= \int\limits^2_0 (8 - \dfrac{3x^2}{2^1}+ \dfrac{3x^4}{2^3}- \dfrac{3x^6}{2x^5}+ \dfrac{3x^8}{2^7} -...) dx

S = 8 \int\limits^2_0dx - \dfrac{3}{2^1} \int\limits^2_0 x^2 dx +  \dfrac{3}{2^3}\int\limits^2_0 x^4 dx -  \dfrac{3}{2^5}\int\limits^2_0 x^6 dx+ \dfrac{3}{2^7}\int\limits^2_0 x^8 dx-...

S = 8(x)^2_0 - \dfrac{3}{2^1*3}(x^3)^2_0 +\dfrac{3}{2^3*5}(x^5)^2_0- \dfrac{3}{2^5*7}(x^7)^2_0+ \dfrac{3}{2^7*9}(x^9)^2_0-...

S= 8(2-0)-\dfrac{1}{2^1}(2^3-0^3)+\dfrac{3}{2^3*5}(2^5-0^5)- \dfrac{3}{2^5*7}(2^7-0^7)+\dfrac{3}{2^7*9}(2^9-0^9)-...

S= 8(2-0)-\dfrac{1}{2^1}(2^3)+\dfrac{3}{2^3*5}(2^5)- \dfrac{3}{2^5*7}(2^7)+\dfrac{3}{2^7*9}(2^9)-...

S = 16-2^2+\dfrac{3}{5}(2^2) -\dfrac{3}{7}(2^2)  + \dfrac{3}{9}(2^2) -...

S = 16-4 + \dfrac{12}{5}- \dfrac{12}{7}+ \dfrac{12}{9}-...

a_0 = 16\\ \\ a_1 = -4 \\ \\ a_2 = \dfrac{12}{5} \\ \\a_3 = - \dfrac{12}{7} \\ \\ a_4 = \dfrac{12}{9}

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3 years ago
Translate to an inequality. Use the variable x.<br> A number is less than 15.
ki77a [65]

Answer:

that's the correct answers

Step-by-step explanation:

thanks for your

3 0
3 years ago
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