The factoring method which can be considered for such a cubic tetranomial expression is; factor by grouping sum of cubes.
<h3>What factoring method can be considered for the polynomial?</h3>
It follows from the task content that the order of the Polynomial is 3 and the polynomial is a tetranomial as it contains 4 terms.
On this note, since 3x³ is not a perfect cube, it follows that the best factorisation method for such a polynomial is; factor by grouping sum of cubes.
Read more on factorisation;
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$10.71
Explanation
Step 1
you can easily solve this by using a rule of three.
Let
x represents the money Eileen will pay if her bag weighs 63 pounds
the proportion is the same
then

Step 2
solve for x

I hope this helps you
Answer:

Step-by-step explanation:
We have the compound inequality:

Let's solve each of them individually first:
We have:

Divide both sides by 2:

Add 1 to both sides:

We have:

Subtract from both sides:

Divide both sides by -4:

Hence, our solution set is:

Answer:
The sign would be positive.
The answer would be 144.
Step-by-step explanation:
Note that there are two negative numbers and two positive numbers:
positive x positive = positive answer.
negative x negative = positive answer.
Solve:
(-4) * (-3) = 12
2 * 6 = 12
12 * 12 = 144
144 would be your answer.
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Answer:
110.5348 minutes
Step-by-step explanation:
The difference from room temperature changes from 103 to 59 in 46 minutes, so that difference can be modeled by the exponential equation ...
Δt = 103×(59/103)^(t/46)
We want to find t for the temperature difference Δt = 91 -64 = 27.
27 = 103×(59/103)^(t/46)
27/103 = (59/103)^(t/46) . . . . . divide by 103
Taking logs gives the linear equation ...
log(27/103) = (t/46)log(59/103)
Multiplying by the inverse of the coefficient of t, we get ...
t = 46·log(27/103)/log(59/103) ≈ 46·(-0.58147346)/(-0.24198521)
≈ 110.5347
It will take about 110.5347 minutes for the turkey to cool to 91 °F internally.
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<em>Comment on 4 decimal places</em>
An answer correct to 4 decimal places (7 significant digits) is a pretty good indication that the problem was worked correctly. However, that level of precision in the timing makes little sense in this context. Most thermometers will take at least a few seconds to register the temperature to within a tenth of a degree or so. This problem is asking for an answer that is within 6 milliseconds and 30 micro-degrees. Neither of these is anywhere near realistic for a kitchen meat thermometer.
More realistic would be an answer to 4 <em>significant figures</em>, a tenth of a minute and a few hundredths of a degree.
(The rate of change at the time of interest is about -0.33 degrees per minute.)