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you just copied and pasted hectictar's answer here!

https://web2.0calc.com/questions/suppose-the-graph-of-y-f-x-includes-the-points-1

Hint:

If the plants are starting at -3, I would say that the plants can grow further down past negative three.
I would draw an open circle at -3, then have the arrow pointing to the left. This shows that the roots can grow lower than -3.

f(1) = 5

f(2) = 3

f(3) = 1

Let  y = f(f(x))

When  x = 2,    y  =  f(f(2))  =  f(3)  =  1

When  x = 3,    y  =  f(f(3))  =  f(1)  =  5

So the points  (2, 1)  and  (3, 5)  must be on the graph of  y = f(f(x))

ab + cd   =   2 * 1  +  3 * 5   =   2 + 15   =   17

Thanks for responding. I don't know what that is or how to use it. Is there another way to do it?

The other roots are 2, -3, and 1 + i.

Let the common difference be \(d\).

Because each term is the sum of the previous term and the common difference, we can write a system of equations: 

\(x + y + d = x - y \ \ \ \ \ \ \ \ \ \ \ \ \ (i)\)

\(x - y + d = x + 3y \ \ \ \ \ \ \ \ \ \ \ (ii)\)

\(x + 3y + d = 1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (iii)\)

From \((i)\), we have: \(y + d = -y\), meaning \(d = -2y\)

Plugging this into \((ii)\) and \((iii)\) gives us a system with 2 equations: 

\(x - y - 2y = x + 3y\)

\(x + 3y + 2y = 1 \)

Solving the system, we find that \(x = 1\) and \(y = 0 \), meaning \(d = 0\), so the fifth term is \(1 + 0 = \color{brown}\boxed{1}\) 

The product of all possible values of a is 14.

The point is (0,-8/3), so the y-coordinate is -8/3.

heureka, Ginger, and Mr.BB, as Ginger calls the guest, had a shot at this

https://web2.0calc.com/questions/help_32678

they all got the same answer, and I looked at their logic

There are 2 cases for number 2: the one digit is a 0, and the one's digit is a 5

For the former case: 1 choice for the one's digit, 5 choices for the 10s digit, and 8 choices for the hundreds digit, for \(8 \times 5 \times 1 = 40\)

For the latter case: 1 choice for the one's digit, 4 choices for the 10s digit (keep in mind we can't re-use 5), and 8 choices for the hundreds digit, for \(1 \times 4 \times 8 = 32\) numbers. 

But, we overcount for 015, 035, 075, and 095, making the total \(40 + 32 - 6 = \color{brown}\boxed{68}\)

Note that it is the same as the previous answer divided by 2. Can you figure out why?

EP's

slope = y1-y2   /   x1-x2

    = 6- -2  /   -2-  -4

    =8 / 2  =4

y=mx+b form

y = 4x + b      Sub in one of the points to find b

6 = 4(-2) + b

b = 14

y = 4x+14      when x = 0     b is the y intercept = 14

you just copied CPhill's answer!!!

I got 72 for the second one but it is in correct
2*5*8 = 80
80-8
72

I'll do number 1, and let you solve for number 2...

For number 1, there are 2 choices for the units digit (0 or 5), 9 choices for the tens digit (anything but the units digit), and 8 choices for the hundreds digit (anything but the tens or unit digit).

This makes for \(2 \times 9 \times 8 = 144\)choices, but we aren't done yet...

Our answer includes numbers like \(085\), which isn't a 3-digit number. 

There is 1 choice for the hundreds digit, 8 choices for the tens digit, and 1 choice for the one's digit, which makes for \(1 \times 8 \times 1 = 8\) numbers. 

So, there are \(144 - 8 = \color{brown}\boxed{136}\) numbers that work

The equation of the first circle is \((x - 2)^2 + (y + 1)^2 = 16\), and the equation of the second circle is \((x - 2)^2 + (y - 5)^2 = 12\)

Subtracting the second equation from the first gives us \((y + 1)^2 - (y - 5)^2 = 4\)

Solving for y, we find that \(y = -{7 \over 3}\). 

Now, plug this into the first equation (either equation works), and solve for x: \((x-2)^2 + ({-4 \over 3})^2 = 16\)

With your x values, calculate the distance between them, and square it...

Good luck!

I don't see 2 variables...

Find the slope of equation 1 as follows: 

\(3x - 4y = 7 \)

\(-4y = 7 - 3x \)

\(4y = -7 + 3x \)

\(4y = 3x - 7\)
\(y = {3 \over 4} x - {7 \over 4}\)

The slope of the line is \({3 \over 4}\), so the slope of the other line is \(-{4 \over 3}\)(negative reciprocal)

Now, convert the second equation into slope-intercept form:

 \(8x + Ay = B\)

\(Ay = B - 8x\)

\(y = {B \over A} - {8 \over A}x\)

\(y = -{8 \over A} x + {B \over A}\)

The slope of this line is \({-8 \over A}\), and we can solve for A: \(-{8 \over A} = -{4 \over 3} \Rightarrow a = 6\)

Now, plugging this back into the equation gives us: \(8x + 6y = B\)

Plug in the given coordinates: \(8 \times 5 + 6 \times 2 = 52\)

So, \(a = 6\) and \(b = 52\), meaning \(a + b = 6 + 52 = \color{brown}\boxed{58}\)

Half of 300 is 150.  Dividing 150 by 1.64, you get 91 (rounded to the nearest integer).  Then divide the other half by 1.32, which gives us 114 (rounded to the nearest integer).  So the answer is 114 - 91 = 23.

3x - 4y= 7

8x + Ay = B

The slope of the first line is 4/3

So....the slope of the second line will be  -3/4 

So  we can write the second line as

Ay  =  -8x + B  ⇒  y  =  (-8/A)x + B/A

So......this implies that   

(-8/A)  = -3/4

A /-8  =  -4/3

A = 32/3 

So we have that

2 =  (-3/4) (5) + B/(32/3)

B = 184/3

So  A + B =   32/3 + 184/3  =  72

Use Simon's Favorite Factoring trick!

ab + 2a + 2b = -16 becomes a(b + 2) + 2b = - 16.

Adding 4 on each side gives us:

a(b + 2) + 2(b + 2) = -12, simplifying to (a + 2)(b + 2) = -12

Doing this factoring method for other equations gives us:

(a + 2)(b + 2) = -12

(a + 2)(c + 2) = 30

(b + 2)(c + 2) = -40.

Multiplying those equations and square rooting gives  us:

(a + 2)(b + 2)(c + 2) = 120

Applying thiss gives us

120/(c + 2) = - 12
120/(b + 2) = 30
120/(a + 2) = - 40

Can you solve for a, b, c from here?

By the British Flag theorem,

DX^2 + 81 = 121 + 100

DX^2 = 140

DX = sqrt(140) = 2sqrt(35)