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$$\displaystyle 3x^3 + x^2 + 5x + 6x^2 + 2x + 10, which \ equals \ 3x^3 + 7x^2 + 7x + 10. The \ answer \ is \ 3 + 1 + 14 + 50, which \ is \ 68.$$

By the triangle inequality  either

80  + 25  >  x               or          25  +  x    >  80

105 > x                                          x  >   80  - 25

x < 105                                          x >   55

Then  the remaining side's shortest length   =   56  inches

See here   :   https://web2.0calc.com/questions/last-two-digits_1

Your not Cphill? lol

Thank you so much, you really helped me a lot

f(z) = z^2 if z is real

f(z) = z + 1 if z is not real

f(i): i+1

f(1): 1^2 = 1

f(-1): -1^2 = -1

f(-i): -i+1

So we got...

(i+1)+1+(-1)+(-i+1) = 2

-Vinculum

Do you think I'm scared guest????

Think again !!!!

-Vinculum

f(x) = 2 - x if x < 1

f(x) = 2x - x^2 if x >= 1

f(-3): 2+3 = 5

f(0): 2-0 = 2

f(3): 2(3)-(3)^2 = -3

5+2-3 = 4

-Vinculum

The answer should be about: yˆ=−0.002x^2−2.051x+31.179

-Vinculum

\(\left(x+2\right)\left(3x^2+x+5\right)\)

\(\mathrm{Distribute\:parentheses}\)

\(x\cdot \:3x^2+xx+x\cdot \:5+2\cdot \:3x^2+2x+2\cdot \:5\)

\(3x^3+7x^2+7x+10\)

Sub these in \(A + B + 2C + 5D\)

\(3+3+2(7)+5(10)\)

\(=70\)

-Vinculum

if we make a list of all numbers which are differences of powers of three, all will be different if we insist that the powers are different. The number of differences with minuend ( a number from which another is subtracted), 3^n is n. The last minuend leaving a difference less than 1000 is 3^6. 1+2+3+4+5+6=21

Note: if you include "0", then you have a total of 22.

If you think you are right

Prove it, guest answerer...

-Vinculum

Is no one going to answer? wow $\displaystyle \int_{dum}^{dum} \ y'all \ dum \ ddum$

$I \ don't \ like \ you \ asinus \ \begin{vmatrix} dum&dum&dumb \\ dum&dum&dumb \\ dum&dum&dumb \end{vmatrix}$

$\displaystyle \begin{pmatrix} please & stop & man \\ I & dont & like \\ This & a & warning \end{pmatrix}$

sorry, that won't be applicable. 

-Cphill

What is c when b=3?

Hello Guest!

\(a:b=c:a\\ a^2=bc\\ a:9=5:a\\ a^2=45\\ bc=45\\ 3c=45\\ c=15\\ \color{blue}c=15\ when\ b=3 \)

  !

Let the total amount of money he started with==M

4/7M == 28, solve for M

4M ==28 x 7

4M == 196

M == 196 / 4 ==$49 - total amount he started with

4/7 * 49 ==$28 - he spent on games

3/7 * 49 ==$21 - he spent on puzzles

28  -  21 ==$7 fewer dollars spent on puzzles.

My computer says this many sums:

        p  n  d  q

1  = (7, 1, 1, 1) == 47
2  = (6, 2, 1, 1) == 51
3  = (5, 3, 1, 1) == 55
4  = (4, 4, 1, 1) == 59
5  = (3, 5, 1, 1) == 63
6  = (2, 6, 1, 1) == 67
7  = (1, 7, 1, 1) == 71
8  = (6, 1, 2, 1) == 56
9  = (5, 2, 2, 1) == 60
10  = (4, 3, 2, 1) == 64
11  = (3, 4, 2, 1) == 68
12  = (2, 5, 2, 1) == 72
13  = (1, 6, 2, 1) == 76
14  = (5, 1, 3, 1) == 65
15  = (4, 2, 3, 1) == 69
16  = (3, 3, 3, 1) == 73
17  = (2, 4, 3, 1) == 77
18  = (1, 5, 3, 1) == 81
19  = (4, 1, 4, 1) == 74
20  = (3, 2, 4, 1) == 78
21  = (2, 3, 4, 1) == 82
22  = (1, 4, 4, 1) == 86
23  = (3, 1, 5, 1) == 83
24  = (2, 2, 5, 1) == 87
25  = (1, 3, 5, 1) == 91
26  = (2, 1, 6, 1) == 92
27  = (1, 2, 6, 1) == 96
28  = (6, 1, 1, 2) == 71
29  = (5, 2, 1, 2) == 75
30  = (4, 3, 1, 2) == 79
31  = (3, 4, 1, 2) == 83
32  = (2, 5, 1, 2) == 87
33  = (1, 6, 1, 2) == 91
34  = (5, 1, 2, 2) == 80
35  = (4, 2, 2, 2) == 84
36  = (3, 3, 2, 2) == 88
37  = (2, 4, 2, 2) == 92
38  = (1, 5, 2, 2) == 96
39  = (4, 1, 3, 2) == 89
40  = (3, 2, 3, 2) == 93
41  = (2, 3, 3, 2) == 97
42  = (3, 1, 4, 2) == 98
43  = (5, 1, 1, 3) == 95
44  = (4, 2, 1, 3) == 99

What is the value of 4^{10} * 8^{20} * 16^4?

Hello Guest!

\({\color{blue}4^{10} \cdot8^{20} \cdot 16^4}=2^{2\cdot 10}\cdot2^{3\cdot 20}\cdot 2^{4\cdot 4}=2^{20+60+16}\color{blue}=2^{96}\)

\(=79228162514264337593543950336\)  !

No Diagram is shown....

-Vinculum

I'm sorry to say this but, think again guest answerer 

\(\mathrm{Apply\:binomial\:theorem}:\quad \left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i\)

(*ahhhh man not this again...)

Just know that:

\(a=2y,\:\:b=-7\)

when we are substituting values in the formula!!!

\(=\sum _{i=0}^5\binom{5}{i}\left(2y\right)^{\left(5-i\right)}\left(-7\right)^i\)

\(=\frac{5!}{0!\left(5-0\right)!}\left(2y\right)^5\left(-7\right)^0+\frac{5!}{1!\left(5-1\right)!}\left(2y\right)^4\left(-7\right)^1+\frac{5!}{2!\left(5-2\right)!}\left(2y\right)^3\left(-7\right)^2+\frac{5!}{3!\left(5-3\right)!}\left(2y\right)^2\left(-7\right)^3+\frac{5!}{4!\left(5-4\right)!}\left(2y\right)^1\left(-7\right)^4+\frac{5!}{5!\left(5-5\right)!}\left(2y\right)^0\left(-7\right)^5\)

\(32y^5-560y^4+3920y^3-13720y^2+24010y-16807\)

The coefficient of y^4 in the expansion of (2y-7)^5 is \(-560\)

-Vinculum

n = d + 30, take it into n + d = 90 =
d + 30 + d = 90

=> = d = 60

        d = 30

So n = 90 - d = 90 - 30 = 60

d = 30

n = 60

60 = 3(30 - x)

30 - x = 20

x = 30 - 20 = 10