Given a polynomial 
and a point 
, we have that
We know that our cubic function is zero at -4, 0 and 5, which means that our polynomial is a multiple of
Since this is already a cubic polynomial (it's the product of 3 polynomials with degree one), we can only adjust a multiplicative factor: our function must be
To fix the correct value for a, we impose 
:
And so we must impose
So, the function we're looking for is
The answer is y=2x+5
To get it use point slope by taking two points and solving .
Slope formula
M= y2-y1/x2-x1
With two points
(0,5)(-5,-5)
M= -5-5/-5-0
M= -10/-5
M=2
2 is slope
Now get one of the points
(0,5) And slope to create equation y=mx+b . Now find b
5=2(0)+b
5=b
So now you can put it all together
Y= 2x+5
Answer:
x(2x−3)(x−3)
Step-by-step explanation:
2x^3−9x^2+9x
= x(2x−3)(x−3)
Try to get the u’s on one side so for this to happen, you would subtract the four from both sides.
u - 4u = 7 - 4
combine like terms
-3u = 3
divide by -3
u = -1
Answer:
i think its the last option.
Step-by-step explanation:
its not the first one because they it can be proved using AAS (but that option isnt available)
You can't prove it with SSS or ASA without more info
AAA doesn't exist
theirfore it must be the last option.
