I do not get it can someone pls help me ill apreciate it

The zeros of a quadratic function are 6 and -4. Which of these choices could be the function?

geniusboy [140]

Answer:

C

Step-by-step explanation:

If x = a is a zero of f(x) then (x - a) is a factor of f(x)

Here the zeros are x = 6 and x = - 4 then the corresponding factors are

(x - 6) and (x - (- 4) ) , that is (x - 6) and (x + 4)

Then f(x) is the product of the factors

f(x) = a(x - 6)(x + 4) ( a is a multiplier )

If a = 1 , then

f(x) = (x - 6)(x + 4) ← is a possible choice → C

6 0

3 years ago

Read 2 more answers

Find the absolute value. |-89= 0 please help thank you

lukranit [14]

the answer is 89

Step-by-step explanation:

it does not matter if the number is negative the absolute value is the number inside the lines

6 0

3 years ago

Read 2 more answers

The solution to a system of linear equations is (-3,-3) Which system of linear equations has this point as its solution? x -5y=-

Ann [662]

Answer:

$x-5y=12$ and $3x+2y=-15$

Step-by-step explanation:

The first system is

$x-5y=-12$ and $3x+2y=-15$

We substitute x=-3 and y=-3

$-3-5*-3=12\ne-12$ and $3*-3+2*-3=-15$

Both equations are not satisfied

The next system is

$x-5y=12$ and $3x+2y=-15$

We substitute x=-3 and y=-3

$-3-5*-3=12$ and $3*-3+2*-3=-15\ne15$

Both equations are satisfied

The next system is

$x-5y=-12$ and $3x+2y=15$

We substitute x=-3 and y=-3

$-3-5*-3=12\ne12$ and $3*-3+2*-3=-15$

Both equations are not satisfied

The next system is

$x-5y=12$ and $3x+2y=15$

We substitute x=-3 and y=-3

$-3-5*-3=12\ne-12$ and $3*-3+2*-3=-15$

Both equations are not satisfied

6 0

4 years ago

Let M be the set of all nxn matrices. Define a relation on won M by A B there exists an invertible matrix P such that A = P BP S

Sophie [7]

Answer:

Recall that a relation is an equivalence relation if and only if is symmetric, reflexive and transitive. In order to simplify the notation we will use A↔B when A is in relation with B.

Reflexive: We need to prove that A↔A. Let us write J for the identity matrix and recall that J is invertible. Notice that $A=J^{-1}AJ$ . Thus, A↔A.

Symmetric: We need to prove that A↔B implies B↔A. As A↔B there exists an invertible matrix P such that $A=P^{-1}BP$ . In this equality we can perform a right multiplication by $P^{-1}$ and obtain $AP^{-1} =P^{-1}B$ . Then, in the obtained equality we perform a left multiplication by P and get $PAP^{-1} =B$ . If we write $Q=P^{-1}$ and $Q^{-1} = P$ we have $B = Q^{-1}AQ$ . Thus, B↔A.

Transitive: We need to prove that A↔B and B↔C implies A↔C. From the fact A↔B we have $A=P^{-1}BP$ and from B↔C we have $B=Q^{-1}CQ$ . Now, if we substitute the last equality into the first one we get