Answer:m∠7
Step-by-step explanation: There isn't much to go off of but from my experience that is the one that is congruent with the other
I hope this helps :)
Answer:
(
g
−
f
)
(
a
)
=
g
(
a
)
−
f
(
a
)
(
g
−f
)
(
a
)
=
g
(
a
)
−
f
(
a
)
=
−
3
a
−
3
−
a
2
−
5
=
−
a
2
−
3a
−
8
Step-by-step explanation:
Answer:
120
Step-by-step explanation:
add 4 and 2 then multipy by 2 and add a zero
Answer:
Octave
Step-by-step explanation:
F to F is an Octave.
Answer:
The quadratic x2 − 5x + 6 factors as (x − 2)(x − 3). Hence the equation x2 − 5x + 6 = 0
has solutions x = 2 and x = 3.
Similarly we can factor the cubic x3 − 6x2 + 11x − 6 as (x − 1)(x − 2)(x − 3), which enables us to show that the solutions of x3 − 6x2 + 11x − 6 = 0 are x = 1, x = 2 or x = 3. In this module we will see how to arrive at this factorisation.
Polynomials in many respects behave like whole numbers or the integers. We can add, subtract and multiply two or more polynomials together to obtain another polynomial. Just as we can divide one whole number by another, producing a quotient and remainder, we can divide one polynomial by another and obtain a quotient and remainder, which are also polynomials.
A quadratic equation of the form ax2 + bx + c has either 0, 1 or 2 solutions, depending on whether the discriminant is negative, zero or positive. The number of solutions of the this equation assisted us in drawing the graph of the quadratic function y = ax2 + bx + c. Similarly, information about the roots of a polynomial equation enables us to give a rough sketch of the corresponding polynomial function.
As well as being intrinsically interesting objects, polynomials have important applications in the real world. One such application to error-correcting codes is discussed in the Appendix to this module.
