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The probability is 16/27.

The equation of the normal at (1,1) is y = 4x - 3.

The product of the roots is \({c \over a} = {n \over 2}\)

Can you take it from here?

There are 44 ways.

There are 28 ways.

The value of N is 50.

Alternatively:

\(81^n=9^{2n}\)

\(9*27*27*9=9*3*9*3*9*9=9^5\)

Hence \(9^{2n}=9^5\)  so \(2n=5\)  or \(n=5/2\)

$83  x  0.70 ==$58.10 what Deon paid for it after 30% discount.

 81^n=9*27*27*9?

n ==Log[9*27*27*9) / log(81)

n==2.5

There are 4 such right triangles:

1 = (5, 13)
2 = (9, 15)
3 = (16, 20)
4 = (35, 37)

The 3rd side of all 4 above is 12, of course.

4-digit permutations==(1, 3, 3, 3),   (2, 2, 3, 3) ==10
5-digit permutations==(1,1,2,3,3), (1, 2, 2, 2, 3), (2, 2, 2, 2, 2)=51

6-digit permutations==(1,1,1,1,3,3), (1,1,1,2,2,3), (1,1,2,2,2,2)=90 

7-digit permutations==(1,1,1,1,1,2,3), (1,1,1,1,2,2,2)==77 

8-digit permutations==(1,1,1,1,1,1,1,3), (1,1,1,1,1,1,2,2)==36 

9-digit permutations==(1,1,1,1,1,1,1,1,2)==9

10-digit permutation ==(1,1,1,1,1,1,1,1,1,1)==1

Grand Total ==10 + 51 + 90 + 77 + 36 + 9 + 1 ==274 permutations.

https://web2.0calc.com/questions/graphing-question_107#r1

Alright, this is gonna be a long one.

first expand this whole thing

\(27t^3+36t^2+27t^2\sqrt{t}+19t\sqrt{t}+12t+3\sqrt{t}+1+\left(\sqrt{t}-3t+1\right)^3\)

\(=27t^3+36t^2+27t^2\sqrt{t}+19t\sqrt{t}+12t+3\sqrt{t}+1-27t^3+18t^2+27t^2\sqrt{t}-17t\sqrt{t}-6t+3\sqrt{t}+1\)

\(=27t^3-27t^3+36t^2+18t^2+27t^2\sqrt{t}+27t^2\sqrt{t}+19t\sqrt{t}-17t\sqrt{t}+12t-6t+3\sqrt{t}+3\sqrt{t}+1+1\) grouping like terms

simplify to get: \(54t^2+54t^2\sqrt{t}+2t\sqrt{t}+6t+6\sqrt{t}+2\)

you technically could group them together in parenthesis and stuff but im not doing that

I already answered this question, and the answer is 1.

Why? Check out my other post on this question by going into my profile and going to answers

Guest, im pretty sure you missed one, which is when m = 2, which would result in 20, and it is valid because the sign is less than or equal to.

b)

Suppose the coefficients are:  (n / r−1), (n / r), (n / r+1). 

Then [n−r+1] / r =8 : 1,
[n−r] / [r+1] =28 : 8, solve for n, r

n ==8  and r ==1
Expand  (x + 1)^8 ==x^8 + 8 x^7 + 28 x^6 + 56 x^5 + 70 x^4 + 56 x^3 + 28 x^2 + 8 x + 1
 

So, the first 3 coefficients are: 1 : 8 : 28

1= 2-3+2+0

2=-2+0x3+2

3=3x2/2+0

4=0x3+2+2

5=2x0+2+3

6=3x2+2x0

7=0+2+2+3

8=3x2+2+0

9=3^2+2x0

10=2^3+2+0

11=2+3^2+0

12=2x3x2+0

14=20-(3x2)

15=20-(3+2)

16=2^3x2+0

18=3^2x2+0

24=2(3x2)+0

32=2^3+2+0

36=3(2x2)+0

64=2^3x2+0

81=3^2+2+0