Answer:
Step-by-step explanation:
<em>Refer to attached</em>
<u>Comparing sales by the ratio of sold to all or</u>
- Sales = sold / (remaining + sold)
<u>Applying to our table:</u>
- New Labels: p/(p+q)
- Old labels: r/(r+s)
Correct answer option is the last one
The ordered pair which is a solution to the given inequality is: C. (2, 1).
<h3>What is an inequality?</h3>
An inequality can be defined as a mathematical relation that compares two (2) or more integers and variables in an equation based on any of the following arguments:
- Less than (<).
- Greater than (>).
- Less than or equal to (≤).
- Greater than or equal to (≥).
Next, we would test the ordered pair with the given inequality to determine a solution as follows:
For ordered pair (4, 4), we have:
3x + 2y < 15
3(4) + 2(4) < 15
12 + 8 < 15
20 < 15 (False).
For ordered pair (3, 3), we have:
3x + 2y < 15
3(3) + 2(3) < 15
9 + 6 < 15
15 < 15 (False).
7x - 4y > 9
7(3) - 4(3) > 9
21 - 12 > 9
9 > 9 (False)
For ordered pair (2, 1), we have:
3x + 2y < 15
3(2) + 2(1) < 15
6 + 2 < 15
8 < 15 (True).
7x - 4y > 9
7(2) - 4(1) > 9
14 - 4 > 9
10 > 9 (True)
For ordered pair (1, 0), we have:
3x + 2y < 15
3(1) + 2(0) < 15
3 + 0 < 15
3 < 15 (True).
7x - 4y > 9
7(1) - 4(0) > 9
7 - 4 > 9
3 > 9 (False)
Read more on inequality here: brainly.com/question/27166555
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Its true for integers if you are using the associative property in addition or multiplication ONLY.
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.
BDE are the correct options