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(6x   +  1 / (3x) )^6

The constant term is

C(6,3)  (6x)³ ( 1 / (3x) )³  =

20 * (6³) (x³) ( 1 / (3³)) ( 1 / x³)  =  

20 * 216 / 27  =  

20 * 8  =  

160

There is no constant.  Do you mean coefficient?

( √x  +   5/x)⁹

The constant term is

C(9, 6) (√x)⁶ (5/x)³    =

84 ( x)³ * 125/ (x³)    =

10500

5!/3!= 20

https://web2.0calc.com/questions/help-please-any-help-would-be-appreciated

shouldn't t be equal to 9, not 7?

Since 2t = 18, then  t = 9

so then:

5t + h = 49

5(9) + h = 49

45 + h = 49

h = 4

x  +  1/x   = 5       multiply   through by  x

x^2  + 1  =  5x

x^2  -  5x  =   -1     complete the square on x

x^2 -   5x  +  25/4  =   -1 + 25/4

(x - 5/2)^2  =   21/4    take the positive root  [ we will get the same result with the negative root ]

x - 5/2  =  √21/2

x =  [ √21  + 5  ] / 2

So   ( x - 2) =   ( √21 + 1 ] / 2

And   ( x - 2)^2  =  [ 21  + 2√21  + 1 ] / 4  =  [ 22 + 2√21] / 4  =  [ 11 + √21] / 2

So

(x - 2)^2    +   25 /( x - 2)^2  =

[ 11 + √21] / 2  +    50/ [ 11 + √21]  =

[rationalize the  denominator of the  second fraction by multiplying  top/bottom  by 11 - √21  ]

[ 11 + √21 ] / 2  +  50 [ 11 - √21]  / [ 121 - 21]  =

[ 11 + √21 ] / 2  +  50 [11 - √21] / 100  =

[ 11 + √21 ] / 2  +  [11 - √21 ]  / 2  =

[11 + 11 ]  / 2  = 

22 / 2   =

11

We need to construct this circle  so that  when the circle  intersects  the line perpendicular to AB  that passes through A, we can label this intersection point, C

So....this makes  AB   =  AC   .....so....two sides of this square will be equal

(2x +  3/x^2)^5  

There  is no constant term

Here is the expansion  :

243/x^10 + 810/x^7 + 32 x^5 + 1080/x^4 + 240 x^2 + 720/x

thank u!

the general term is

\(T_n=\binom {9}{n} \left(\frac{5}{x}\right)^n(x^{0.5})^{9-n}\\ T_n=\binom {9}{n} (5^n)(x^{-n})(x^{4.5-0.5n})\\ T_n=\binom {9}{n} (5^n)(x^{4.5-1.5n})\\ \text{for the constant term} \quad \\ 4.5-1.5n=0\\ 9-3n=0\\ 3n=9\\ n=3\\ T_3=\binom {9}{3} (5^3)\\ T_3=84 *125\\ T_3=10500 \)

You need to check this.

Coding:

T_n=\binom {9}{n} \left(\frac{5}{x}\right)^n(x^{0.5})^{9-n}\\
T_n=\binom {9}{n} (5^n)(x^{-n})(x^{4.5-0.5n})\\
T_n=\binom {9}{n} (5^n)(x^{4.5-1.5n})\\
\text{for the constant term}  \quad \\
4.5-1.5n=0\\
9-3n=0\\
3n=9\\
n=3\\
T_3=\binom {9}{3} (5^3)\\
T_3=84 *125\\
T_3=10500
 

Sorry.... from P to H is   3/10 of the total and this = 12

   T ate 3/10 of the total.....so 12     Yes....correct...ignore my first answer above...  

P  T  H

2+3+5 = 10

from 2 to 5  is   2 /10 of the total  and this equals 12   

   T  ate 3/10 of the total   which is   18 

WolframAlpha.com  says 10500    but I do not know how to get there !     Let's see what Melody has to say.....

thanks

Let the square side AB = 2

Let's denote the midpoints of a side AB with a letter M, and a side CD with a letter Q

MP = tan(15°) * 1            MP = 0.268   

PQ = BC - MP                 PQ = 1.732   

 ∠DCP => q                    tan(q) = PQ / [(AB)/2]      q = 60°    

why? pretty sure that's incorrect

\(z\bar{z}^3+\bar{z}z^3=350\\ (a+bi)(a-bi)^3+(a-bi)(a+bi)^3=350\\ (a^2-(bi)^2)[(a-bi)^2+(a+bi)^2]=350\\ (a^2+b^2)[(a^2-2abi-b^2)+(a^2+2abi-b^2)]=350\\ (a^2+b^2)[(2a^2-2b^2)]=350\\ (a^2+b^2)(a^2-b^2)=175\\ a^4-b^4=175 \)

So the solution are   (

\(z=+4\pm3i\\ z=-4\pm3i\)

So assuming i have made no silly mistakes the area is 48 units squared

coding:

z\bar{z}^3+\bar{z}z^3=350\\
(a+bi)(a-bi)^3+(a-bi)(a+bi)^3=350\\
(a^2-(bi)^2)[(a-bi)^2+(a+bi)^2]=350\\
(a^2+b^2)[(a^2-2abi-b^2)+(a^2+2abi-b^2)]=350\\
(a^2+b^2)[(2a^2-2b^2)]=350\\
(a^2+b^2)(a^2-b^2)=175\\
a^4-b^4=175

The constant term is C(9,2)*5^4 = 22500.

answer said The constant term is 0. We might say there is "no constant term" in such a situation, but this is a colloquial way of saying that the constant term is equal to 0. 

15*6*8*3!

Let the number of hours when they will be equal =H

25 + 15H =20H

20H - 15H =25

5H =25 divide both sides by 5

H =25/5

H = 5 hours-when they will be equal.

ok, next time give your reason right from the beginining.

Don't vote people down. Thank them for their interest in you question, explain why you think they are incorrect.

Then maybe you will learn together when someone else answers.

If you are both members you could even become friends.

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If you are not sure maybe one of our other users can help you find out. 

I can't, I do not know about such things. 

But I am sure others would be able to .