breadstickim01

(0 , 7) 

(a + ab) + (a - ab) ==> 250 + (-240)

2a ==> 10

a=5

solve it from here by plugging in a=5 into the equation. 

1. first we simplfy 

y = -5x^2 -6x + 10 is the most simplfied version 

* We can now convert this to vertex form

y= -5 ( x^2 + 6/5 x - 2 ) 

==> y = -5 ( x + 3/5)^2 +59/5  is the vertex form

vertex ==> (h,k)  and the standard vertex form == >  y = a ( x − h ) 2 + k

thus vertex ==> (-3/5 , 59/5)

for a quadratic to have a double root we want the variables of b^2 -4ac = 0

we can combine like terms in this quadratic via simplification and get 

x^2+11x+4x + k 

==> x^2 + 15x + k = 0

now we just plug this into the discriminant equation ( b^2 - 4ac = 0  ) 

== > 15^2 - 4 x 1 x c =0

simplfy == > 225 - 4c = 0 

thus c = 56.25

please check if this is right 

simply : x^2-9x-52 = 0 

than we can plug this in to the quadratic equation : 

than we can get the roots to equation : 13 and -4 

I think that we should start with factoring/finding a quadratic formula that we know has a root of 5 or -5 

we can get a factored version of a quadratic formula with roots of 5 and -5  ==> (x-5)(x+5) = we can get x^2-25 if we simplefy it 

so your anwser would be 

x^2 - 0x + (-25) 

thanks 

add -3x^2 to both sides 

add 9x to both sides 

add 4 to both sides

10x^2+10x-1 = 0 

we can plug into the quadratic formula : 

- we will get (-5+sqrt 35) / 10.  or (-5-sqrt 35) / 10

then you can plug these value into the other expression we can get 412 / 199

i think you should work on getting to know the quadratice equation more + simplfying equations in general

subtract 4x^2 from both sides 

- get 1x^2 

-add 17x on both sides = get 6x on the left side 

-subtract 6 from both sides = -2 on the right side 

if you do this you SHOULD get , 

x^2 + 6x - 2 = 0 

plug into quadratic formula (search up) : where ax^2 + bx + c is the formate ( a = 1.  b = 6  c = -2 )
we should get the solution: -3+sqrt(11) or -3-sqrt(11) and they represent a and b 

1/ (-3+sqrt(11))^3 + 1/ (-3-sqrt(11))^3 = 63/2 or 31.5

see if you can get it!

and someone please check my work

Srry mb gang 

a/3 + 1 = (( a+3) - (a^2 + 2))/a

a/3 + 1 =( a^2 - a - 1 ) / a 

a ( a/3 + 1 ) = a^2 - a - 1 

a^2/3 + a = a^2 - a -1 

a^2/3 = a^2 - 2a -1 

0 = (2a^2 - 2a) / 3  - 1 

then we can plug this equation into the quadratic equation, 

this gives us the answer 

(3+sqrt 15 )/2 or (3-sqrt15)/2 

please if anyone could check my answer that would be a fire 

(-24/5 ,  0 ) ? 

For the first one assuming that f(1) = 20-5 then im guessing the function is that when n increases the number of fives subtracted from 20 increases via the number of n (f(n)).

f(10) = -20-5-5-5-5-5-5-5-5-5-5 sooooo it has 5 tens 

4. f(1) =  - 6 as its a recursive function meaning you need to find the previous function number