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The product of the numbers is 12.

y = 11/3.

We all know you're lying.

this isn't A O P S, it's from my school teacher. He said he found it online.

This is not AoPS, my school teacher posted as summer assignment. I don't know if he found it online or not.

Do not cheat on A o P S homework.

b + c = -28

\((c_1, c_2, d_1, d_2) = (-4/5, 3/5, 4/5, -3/5)\)

Using the python script
list=[]
for i in range(0,151):
    x=i*8800
    if x % 10000 == 0:
        list.append(i)
print(list)

The answers are 0, 25, 50, 75, 100, 125, and 150

You might find it easier if you swap the z for an x and the swap it back at the end

\((4-z)(z+7) ≥ 2\\ (4-x)(x+7) ≥ 2\\ -x^2 -3x +28 \ge 2\\ -x^2 -3x +26 \ge 0\\ x^2+3x -26 \le 0\\ \)

 \(y=x^2+3x-26\\ \text{Is a concave up parabola}\)

So it will be less than 0 between the roots.  Draw it to see what i mean

Use the quadratic formula to find the roots

etc

Note that if three numbers x, y, z form an arithmetic progression, then \(y = \dfrac{x + z}2\).

Also, if three numbers a, b, c form a geometric progression, then \(b^2 = ac\).

Therefore, \(4^2 = 2\cdot x\). Solving gives x = 8.

Can you continue from here?

In the coordinate plane, let R be the region defined by 2 <= |x| and |y| <= 4. Find the area of R.

Hello Guest!

R = 4x  |  x >= 2

  !

2/3 of the way from 6 to 12 would be 10.  Agreed?

2/3(12-6) = 2/3 (6) = 4          6+4=10      That is how to find it more formally.

so

Given R(-6,19)  and S(15,-5), what are the coordinates of the point on segment RS two-thirds of the distance from R to S?

the x value will be    -6  +  2/3(15 - -6) 

and you can find the y value

You are expected to set up the inequality and then solve it.

What have you already done to at least start this problem?

You can make your own collection of troll posts if you want to Elijah.

Just start a new thread called 'Elijah's collection of troll posts" 

This will be saved in your own Watchlist.

Then you can slowly add posts with the address of the threads that you want to save.  :)

\(99^0*99^1*99^2* .....*99^{99}\\ =99^{0+1+2+3+4+...99}\\ =99^{\text{an even number}}\)

99^1=99

99^2=9801

99^3=ends in 99

99^4 ... ends in 01

etc

99 ^ (any even positive integer) will end in 01

Hi Guest!

Let the first term of the arithmetic progression be \(a_0\)

Let the common difference between two successive terms be \(d\)

Then,  \(a_n=a_0+d(n-1)\)

The third term is when n=3:

\(a_3=a_0+2d\)

We are given that the third term is 94:

\(94=a_0+2d\)               (Equation 1)

The 6th term is when n=6:

\(a_6=a_0+5d\)

Given that the 6th term is 85:

\(85=a_0+5d \)           (Equation 2)

Subtract Equation 2 from Equation 1:

\(94-85=a_0+2d-a_0-5d \iff 9 = -3d \iff d=-3\)

Substituting in Equation 1:

\(94=a_0+2(-3) \iff a_0=100\)

The 15th term is when n=15:

\(a_{15}=a_0+14d=100+14(-3)=100-42=58\)

Hope this helped!

\(4x^2+5x+2-x^2+x+6 = 3x^2+6x+8\)

3(x^2+2x+(8/3))

3(x^2+2x+1-1+8/3)

3((x+1)^(2)-1+8/3)

3(x+1)^(2) + 5

-Vinculum

The numbers 1,b,c form an arithmetic progression and the numbers 1,b,c+1 form a geometric progression. Find c.  

arithmetic  --------     1, 2, 3 

geometric  --------     1, 2, 4         so c would equal 3  

_.

\(f(f(f(5)))=\)\(\frac{\left(\frac{\left(\frac{\left(x-1\right)}{\left(x+1\right)}-1\right)}{\left(\frac{\left(x-1\right)}{\left(x+1\right)}+1\right)}-1\right)}{\left(\frac{\left(\frac{\left(x-1\right)}{\left(x+1\right)}-1\right)}{\left(\frac{\left(x-1\right)}{\left(x+1\right)}+1\right)}+1\right)}\)

That simplifies to \(\frac{-x-1}{x-1}\)

Substituting 5 gives \(\frac{-\left(5\right)-1}{\left(5\right)-1}\)

=\(-3/2\)

So the guest in 3rd post was right

\(\displaystyle \prod_{n=0}^{99} 99^n\)

\(99^{4950}\)

As the guest before mentioned the last 2 digits are 01

-Vinculum

Hello Guest, this question has been answered in 2 other pages here:

Take a look(both have the same answers)!!

https://web2.0calc.com/questions/help-geometry_86109

https://web2.0calc.com/questions/diameter-of-a-right-cylindrical

-Vinculum

Hey this question has already been answered by CPhill

Take a look: https://web2.0calc.com/questions/diameter-of-a-right-cylindrical

-Vinculum