See explanation below.
Explanation:
The 'difference between roots and factors of an equation' is not a straightforward question. Let's define both to establish the link between the two..
Assume we have some function of a single variable
x
;
we'll call this
f
(
x
)
Then we can form an equation:
f
(
x
)
=
0
Then the "roots" of this equation are all the values of
x
that satisfy that equation. Remember that these values may be real and/or imaginary.
Now, up to this point we have not assumed anything about
f
x
)
. To consider factors, we now need to assume that
f
(
x
)
=
g
(
x
)
⋅
h
(
x
)
.
That is that
f
(
x
)
factorises into some functions
g
(
x
)
×
h
(
x
)
If we recall our equation:
f
(
x
)
=
0
Then we can now say that either
g
(
x
)
=
0
or
h
(
x
)
=
0
.. and thus show the link between the roots and factors of an equation.
[NB: A simple example of these general principles would be where
f
(
x
)
is a quadratic function that factorises into two linear factors.
48.. i don’t have the work but hopefully someone else does
If it's add you subtract if it's subtract you add
Answer:
I was unsure about the way your question was written so I solved for two equations: f(x)=14(5-x)*2 and f(x)=14(5-x)+2
Step-by-step explanation:
Answer:
The x-coordinate of the point changing at ¼cm/s
Step-by-step explanation:
Given
y = √(3 + x³)
Point (1,2)
Increment Rate = dy/dt = 3cm/s
To calculate how fast is the x-coordinate of the point changing at that instant?
First, we calculate dy/dx
if y = √(3 + x³)
dy/dx = 3x²/(2√(3 + x³))
At (x,y) = (1,2)
dy/dx = 3(1)²/(2√(3 + 1³))
dy/dx = 3/2√4
dy/dx = 3/(2*2)
dy/dx = ¾
Then we calculate dx/dt
dx/dt = dy/dt ÷ dy/dx
Where dy/dx = ¾ and dy/dt = 3
dx/dt = ¾ ÷ 3
dx/dt = ¾ * ⅓
dx/dt = ¼cm/s
The x-coordinate of the point changing at ¼cm/s