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anastassius [24]
4 years ago
11

Which function has only one x-intercept at (-6, 0)?

Mathematics
2 answers:
Lelu [443]4 years ago
6 0

Answer:

<h2>The function f(x) = (x - 6)(x - 6) has only one x-intercept. But at (6, 0) not at (-6, 0).</h2>

Step-by-step explanation:

The intercept form of a quadratic equation (parabola):

y=a(x-p)(x-q)

p, q - x-intercepts

Therefore

The function f(x) = x(x - 6) = (x - 0)(x - 6) has two x-intercepts at (0, 0) and (6, 0)

The function f(x) = (x - 6)(x - 6) has only one x-intercept at (6, 0)

The function f(x) = (x + 6)(x - 6) = (x - (-6))(x - 6)

has two x-intercept at (-6, 0) and (6, 0)

The function f(x) = (x + 1)(x + 6) = (x - (-1))(x - (-6))

has two x-intercepts at (-1, 0) and (-6, 0).

shusha [124]4 years ago
3 0

Answer:

the answer is d on edge

Step-by-step explanation:

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Complete the proof....
abruzzese [7]

Answer:

The correct options are;

1. Definition of supplementary angles

2. m∠1 + m∠2 = m∠1 + m∠3

3. m∠2 = m∠3

4. Definition of Congruent Angles

Step-by-step explanation:

The two column proof is presented as follows;

Statement           {}                                  Reason

1. ∠1 and ∠2 are supplementary    {}     Given

 {} ∠1 and ∠3 are supplementary

2. m∠1 + m∠2 = 180°        {}                    Definition of supplementary angles

{}   m∠1 + m∠3 = 180°

3. m∠1 + m∠2 = m∠1 + m∠3      {}          Transitive Property

4. m∠2 = m∠3       {}                               Subtraction Property of Equality

5. ∠2 ≅ ∠3       {}                                    Definition of Congruent Angles

Given that angles ∠1 and ∠2 are supplementary angles and angles ∠1 and ∠3 are are also supplementary angles, then the sums of m∠1 + m∠2 and m∠1 + m∠3 are equal, therefore, ∠2 and ∠3 have equal quantitative value and therefore ∠2 = ∠3 and by definition, ∠2 ≅ ∠3.

3 0
3 years ago
Solve for X 4x−1 =3 Answer:
Verdich [7]

Add 1 to both sides of the equation to get 4x = 4.

Divide both sides by 4 and <em>x = 1</em>.

7 0
3 years ago
Read 2 more answers
A line segment or ray that is perpendicular to the segment at its midpoint is called
Dimas [21]
Perpendicular bisector
3 0
3 years ago
I need help on these two.
Naily [24]
The answer to the first question is 30.

8 0
4 years ago
Una recta con pendiente -2 ´pasa por el punto P(5, -1). La abscisa del punto Q que está en esa recta es 1. Encuentre la ordenada
oksian1 [2.3K]

Answer:

La ordenada de Q es 7.

Step-by-step explanation:

La ecuación de una recta es:

y = a*x + b

donde a es la pendiente y b es la ordenada al origen.

Para este problema, sabemos que la pendiente es -2

Entonces:

a = -2

y = -2*x + b

También sabemos que esta recta pasa por el punto (5, -1)

Esto significa que cuando evaluamos la ecuación en x = 5, y toma el valor y = -1

Si reemplazamos esos valores en la ecuación, obtenemos:

-1 = -2*5 + b

-1 = -10 + b

-1 + 10 = b

9 = b

Entonces la ecuación de la recta es:

y = -2*x + 9

Sabemos que el punto Q está en la línea, y la abscisa de este punto es 1.

Entonces este punto es Q (1, k)

donde k es la ordenada de este punto.

Porque nuestra línea también pasa por este punto, podemos concluir que cuando x = 1, y = k

Reemplazando eso en la ecuación tenemos:

k = -2*1 + 9 = -2 + 9 = 7

k = 7

La ordenada de Q es 7.

7 0
3 years ago
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