Answer:
<h2>
y= -x -10</h2>
y=mx+b
plug in -6 for y and -4 for x and - for m
-6=-(-4)+b
solve for b
b=-10
set up equation
Factor 4
4=1 times 4
2 times 2
they don't add to 2
set up equation
x+y=2
xy=4
first equation, subtract x from both sides
y=2-x
subsitute for y
x(2-x)=4
distribute
2x-x^2=4
add x^2
2x=x^2+4
subtract 2x
0=x^2-2x+4
use quadratic formula which is
if you have ax^2+bx+c=0 then
x=

so
1x^2-2x+4=0
a=1
b=-2
c=4
x=

x=

x=

we have

and that doesn't give a real solution
therefor there are no real solutions
but if you want to solve fully
x=

i=

x=

x=

x=

or x=

(those are the 2 numbers)
Let d = 0.825252525..., then

Therefore,
Answer:
f(n) = -n^2 -3n +5
Step-by-step explanation:
Suppose the formula is ...
f(n) = an^2 +bn +c
Then we have ...
f(1) = 1 = a(1^2) +b(1) +c
f(2) = -5 = a(2^2) +b(2) +c
f(3) = -13 = a(3^2) +b(3) +c
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Here's a way to solve these equations.
Subtract the first equation from the second:
-6 = 3a +b . . . . . 4th equation
Subtract the second equation from the third:
-8 = 5a +b . . . . . 5th equation
Subtract the fourth equation from the fifth:
-2 = 2a
a = -1
Then substituting into the 4th equation to find b, we have ...
-6 = 3(-1) +b
-3 = b
and ...
1 = -1 +(-3) +c . . . . . substituting "a" and "b" into the first equation
5 = c
The formula is ...
f(n) = -n^2 -3n +5
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