A perfect square trinomial can be represented by a square model with equivalent length and width. Which polynomial can be repres
2 answers:
The correct answer is
In fact, this is a trinomial of the form
, whose solutions are given by
Using this formula for the trinomial of the problem, we find:
<span>we see that this trinomial has two coincident solutions (x=3 with multiplicity 2). This means that it can be rewritten as a perfect square, in the following form:
</span>
<span>
</span>
In fact, this is a trinomial of the form
Using this formula for the trinomial of the problem, we find:
<span>we see that this trinomial has two coincident solutions (x=3 with multiplicity 2). This means that it can be rewritten as a perfect square, in the following form:
</span>
</span>
Answer:
a. x2 – 6x + 9
Step-by-step explanation:
In order to solve this you just have to try and factorize the options and the result should be two exact binomials. Remember that the formula for perfect square trinomial is:
So we only have to factorize x2-6x+9=
As you can see, a is equal to X, and b equals 3:
(x-3)(x-3)=x2-6x+9
SO this is a perfect square trinomial.
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Let 
be the number of hours that the waitress works in the evening.
We know for our problem that she works
in the afternoon and that she works <span>1.75 hours less in the afternoon than in the evening, so:
</span>
<span>Since </span>
, we can rewrite our expression:
We can conclude that she works 3.375 hours in the evening, or expressed as a mixed fraction:
hours.
We know for our problem that she works
</span>
<span>Since </span>
We can conclude that she works 3.375 hours in the evening, or expressed as a mixed fraction:
Initial repair cost = $1200
Gratuity = 15% of the initial repair cost
= 15% of 1200
= 15 × 12
= 180
Hence, the gratuity for the service is $180.
Answer:
We want a polynomial of smallest degree with rational coefficients with zeros in ,
and -3. The last root gives us the factor (x+3). Hence, our polynomial is
where is a polynomial with rational coefficients and roots
and
. The root
gives us a factor
, but in order to obtain rational coefficients we must consider the factor
.
An analogue idea works with . For convenience write
. This gives the factor
. Hence,
Notice that . So, in order to satisfy the last condition we divide by 3 the whole polynomial, without altering its roots. Finally, the wanted polynomial is
Step-by-step explanation:
We must have present that any polynomial it's determined by its roots up to a constant factor. But here we have irrational ones, in order to eliminate the irrational coefficients that a factor of the type will introduce in the expression, we need to multiply by its conjugate
. Hence, we will obtain
that have rational coefficients. Finally, the last condition is given with the intention to fix the constant factor. Usually it is enough to evaluate in the point and obtain the necessary factor.