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Deffense [45]
3 years ago
14

2. Is the following quadratic equation in Vertex form? y = - (x + 2) + 4 O NO O Yes

Mathematics
1 answer:
puteri [66]3 years ago
3 0

Answer:

Step-by-step explanation:

No.  y = - (x + 2) + 4 is not even a quadratic.

If you want a quadratic you must indicate squaring:  y = - (x + 2)^2 + 4.

This IS in vertex form; the vertex is at (-2, 4).

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Step-by-step explanation:

+ and - cancel. We have 5 + and 3 - so that leaves us with 2 + so the answer is +2.

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Step-by-step explanation:

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3 years ago
Just the answer please
nordsb [41]

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3 years ago
Which is a correct two column proof. Given &lt;4 and
vladimir1956 [14]

The initial statement is:    QS = SU   (1)

                                    QR = TU    (2)

 

We have to probe that:  RS = ST

 

 

Take the expression (1):                     QS       =   SU

We multiply both sides by R                (QS)R   =   (SU)R

 

 

But    (QS)R = S(QR)     Then:            S(QR)   =   (SU)R     (3)

 

From the expression (2):  QR = TU. Then, substituting it in to expression (3):

 

                                                       S(TU)   =   (SU)R     (4)

 

But  S(TU) = (ST)U  and (SU)R = (RS)U

 

Then, the expression (4) can be re-written as:

 

                                                      (ST)U    =    (RS)U

 

Eliminating U from both sides you have:     (ST) = (RS)    The proof is done.

4 0
3 years ago
Determine the graph of the polar equation r=4/(1-1/2cosx(theta))
crimeas [40]

Answer:

  see below

Step-by-step explanation:

We assume you want the graph of ...

  r=\dfrac{4}{1-\frac{1}{2}\cos{(\theta)}}

A graphing calculator or spreadsheet is useful for this.

__

You know cos(θ) = cos(-θ), so the graph is symmetrical about the x-axis. You can evaluate the function at a few points to find the general outline.

  r at 0° = 8

  r at 30° ≈ 7.05

  r at 45° ≈ 6.19

  r at 60° ≈ 5.33

  r at 90° = 4

  r at 120° = 3.2

  r at 135° ≈ 2.96

  r at 150° ≈ 2.79

  r at 180° ≈ 2.67

6 0
3 years ago
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