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Kay [80]
3 years ago
11

I) Find the first 4 terms in the expansion of

Mathematics
1 answer:
Katyanochek1 [597]3 years ago
6 0

Answer:

(i)  64+192 x^2+240 x^4+160x^6

(ii) 1072

Step-by-step explanation:

<u>Part (i)</u>

Using Binomial series formula:

(2+x^2)^6=2^6+6C1 \cdot 2^{6-1}\cdot x^2+6C2 \cdot 2^{6-2}\cdot(x^2)^2+6C3 \cdot 2^{6-3}\cdot(x^2)^3

              =64+6 \cdot 32\cdot x^2+15 \cdot 16\cdot x^4+20 \cdot 8 \cdot x^6

              =64+192 x^2+240 x^4+160x^6

<u>Part (ii)</u>

<u />(1-\frac{3}{x^2})^2=(1-\frac{3}{x^2})(1-\frac{3}{x^2})

               =1-\dfrac{6}{x^2}+\dfrac{9}{x^4}

(2+x^2)^6(1-\frac{3}{x^2})^2=(64+192 x^2+240 x^4+160x^6)\left(1-\dfrac{6}{x^2}+\dfrac{9}{x^4} \right)

=64-\dfrac{384}{x^2}+\dfrac{576}{x^4}+192x^2-1152+\dfrac{1728}{x^2}+240x^4-1440x^2+2160+...

There is no need to keep expanding since the remaining will include the variable.

Therefore, the term independent of x = 64 - 1152 + 2160 = 1072

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Factor completely<br> B. (x+5)²-2(x+5)-24<br> C. (x+2)²-8(x+2)+15<br> D. 2(2x-1)²-9(2x-1)+4
Sladkaya [172]
B.
{x}^{2} + 10x + 25 - 2x - 10 - 24 = \\ x {}^{2} + 8x - 9 = \\ (x - 1)(x + 9)

C.
{x}^{2} + 4x +4 - 8x - 16 + 15 = \\ x {}^{2} - 4x + 3 = \\ (x - 3)(x - 1)
D.
2( {4x}^{2} - 4x + 1) - 18x + 9 + 4 = \\ 8x {}^{2} - 8x + 2 - 18x + 13 = \\ 8x {}^{2} - 26x + 15

please mark as brainliest.
3 0
3 years ago
Simplify u^2+3u/u^2-9<br> A.u/u-3, =/ -3, and u=/3<br> B. u/u-3, u=/-3
VashaNatasha [74]
  The correct answer is:  Answer choice:  [A]:
__________________________________________________________
→  "\frac{u}{u-3} " ;  " { u \neq ± 3 } " ; 

          →  or, write as:  " u / (u − 3) " ;  {" u ≠ 3 "}  AND:  {" u ≠ -3 "} ; 
__________________________________________________________
Explanation:
__________________________________________________________
 We are asked to simplify:
  
  \frac{(u^2+3u)}{(u^2-9)} ;  


Note that the "numerator" —which is:  "(u² + 3u)" — can be factored into:
                                                      →  " u(u + 3) " ;

And that the "denominator" —which is:  "(u² − 9)" — can be factored into:
                                                      →   "(u − 3) (u + 3)" ;
___________________________________________________________
Let us rewrite as:
___________________________________________________________

→    \frac{u(u+3)}{(u-3)(u+3)}  ;

___________________________________________________________

→  We can simplify by "canceling out" BOTH the "(u + 3)" values; in BOTH the "numerator" AND the "denominator" ;  since:

" \frac{(u+3)}{(u+3)} = 1 "  ;

→  And we have:
_________________________________________________________

→  " \frac{u}{u-3} " ;   that is:  " u / (u − 3) " ;  { u\neq 3 } .
                                                                                and:  { u\neq-3 } .

→ which is:  "Answer choice:  [A] " .
_________________________________________________________

NOTE:  The "denominator" cannot equal "0" ; since one cannot "divide by "0" ; 

and if the denominator is "(u − 3)" ;  the denominator equals "0" when "u = -3" ;  as such:

"u\neq3" ; 

→ Note:  To solve:  "u + 3 = 0" ; 

 Subtract "3" from each side of the equation; 

                       →  " u + 3 − 3 = 0 − 3 " ; 

                       → u =  -3 (when the "denominator" equals "0") ; 
 
                       → As such:  " u \neq -3 " ; 

Furthermore, consider the initial (unsimplified) given expression:

→  \frac{(u^2+3u)}{(u^2-9)} ;  

Note:  The denominator is:  "(u²  − 9)" . 

The "denominator" cannot be "0" ; because one cannot "divide" by "0" ; 

As such, solve for values of "u" when the "denominator" equals "0" ; that is:
_______________________________________________________ 

→  " u² − 9 = 0 " ; 

 →  Add "9" to each side of the equation ; 

 →  u² − 9 + 9 = 0 + 9 ; 

 →  u² = 9 ; 

Take the square root of each side of the equation; 
 to isolate "u" on one side of the equation; & to solve for ALL VALUES of "u" ; 

→ √(u²) = √9 ; 

→ | u | = 3 ; 

→  " u = 3" ; AND;  "u = -3 " ; 

We already have:  "u = -3" (a value at which the "denominator equals "0") ; 

We now have "u = 3" ; as a value at which the "denominator equals "0"); 

→ As such: " u\neq 3" ; "u \neq -3 " ;  

or, write as:  " { u \neq ± 3 } " .

_________________________________________________________
6 0
3 years ago
Help me please
Viktor [21]
(3x+4) units.

Working;
AM=AD+DM
      =x+5+2x-1
      =(3x+4) units

8 0
3 years ago
A car travels 360 miles in a period of 180 minutes. Find the average velocity of the car in miles per hour over this time period
Serggg [28]
The simple formula for velocity is v=d/t.  Velocity equals distance divided by time.  360 miles divided by 180 minutes.  360/180=2.  The car travels at 2 miles per MINUTE.

Now, to change to miles per HOUR.  There are 60 minutes in an hour, so 2*60=120 miles per hour.

Answer: 120 mph
4 0
4 years ago
Read 2 more answers
State how many solutions, 3 + 2x + 1 = 2x + 3
AlexFokin [52]

Answer:

Step-by-step explanation:

3 + 2x + 1 = 2x + 3

2x + 4 = 2x + 3

4 ≠ 3

no solution

5 0
4 years ago
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