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Elina [12.6K]
3 years ago
14

Francisco and Ryan are stuck simplifying radical expressions. Francisco has simplified to the quantity of x to the one half powe

r, over x to the three eighteenth power. Ryan has simplified to the twenty seventh root of the quantity of x to the second times x to the third times x to the fourth. Using full sentences, describe how to fully simplify Francisco and Ryan's expressions. Describe if Francisco and Ryan started with equivalent expressions or if they started with expressions that are not equal.
Mathematics
1 answer:
Wewaii [24]3 years ago
4 0
X to the one half power, over x to the three eighteenth power is equal to x to the one half power, divided by x to the one sixth power, which equals x to the power of (one half minus one sixth), or x to the one third power. 

<span>The twenty seventh root of the quantity of x to the second times x to the third times x to the fourth equals the twenty seventh root of x to the ninth power which equals x to the one third power. </span>

<span>I cannot say whether Francisco and Ryan started with equivalent expressions but on final simplification they ended up the same. One might reasonably assume they started with equivalent expressions, but who knows if they made any mistakes in their simplifications.</span>
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Answer:

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Step-by-step explanation:

The reasoning is that in the "IRL Equation" the polls showed that orange juice was prefered over milk, and didn't matter whether they wanted a bagel or muffin. All that matters in this equation is that it is one sided (For the 1st poll that is) and now you are able to narrow down the possibilities of the answer(s).

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4 years ago
Deviation is the difference between any value and the mean of the set.
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7 0
3 years ago
Please help me answer this question
avanturin [10]

By <em>direct</em> substitution and simplification, the <em>trigonometric</em> function z = cos (2 · x + 3 · y) represents a solution of the <em>partial differential</em> equation  \frac{\partial^{2} t}{\partial x^{2}} - \frac{\partial^{2} t}{\partial y^{2}} = 5\cdot z.

<h3>How to analyze a differential equation</h3>

<em>Differential</em> equations are expressions that involve derivatives. In this question we must prove that a given expression is a solution of a <em>differential</em> equation, that is, substituting the variables and see if the equivalence is conserved.

If we know that z = \cos (2\cdot x + 3\cdot y) and \frac{\partial^{2} t}{\partial x^{2}} - \frac{\partial^{2} t}{\partial y^{2}} = 5\cdot z, then we conclude that:

\frac{\partial t}{\partial x} = -2\cdot \sin (2\cdot x + 3\cdot y)

\frac{\partial^{2} t}{\partial x^{2}} = - 4 \cdot \cos (2\cdot x + 3\cdot y)

\frac{\partial t}{\partial y} = - 3 \cdot \sin (2\cdot x + 3\cdot y)

\frac{\partial^{2} t}{\partial y^{2}} = - 9 \cdot \cos (2\cdot x + 3\cdot y)

- 4\cdot \cos (2\cdot x + 3\cdot y) + 9\cdot \cos (2\cdot x + 3\cdot y) = 5 \cdot \cos (2\cdot x + 3\cdot y) = 5\cdot z

By <em>direct</em> substitution and simplification, the <em>trigonometric</em> function z = cos (2 · x + 3 · y) represents a solution of the <em>partial differential</em> equation  \frac{\partial^{2} t}{\partial x^{2}} - \frac{\partial^{2} t}{\partial y^{2}} = 5\cdot z.

To learn more on differential equations: brainly.com/question/14620493

#SPJ1

3 0
2 years ago
Please help me do the substitution for this problem and the check thank you
Alexxandr [17]
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8 0
3 years ago
(40 POINTS) PLZ HELP ASAP! DOMAIN AND RANGE! CLICK LINK
MAVERICK [17]

For finding the values, we look at the x - value given. Then we move to where it is on the graph and find its y  value.

In part A, when x = -4, the y value looks to be between -3 and -4. Let's put it in the middle and estimate it at -3.5

In part B, when x = 1, the process is similar.  Go to x = 1, then go to the graph, then go to the y value.  That looks to be at y = -3.

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We answered Part D when observing part C. It's NOT a function because there is not EXACTLY ONE value of y for an x.

And finally, because it's not a function - finding the domain and range is a waste of time. You can't find the domain of something that's not a function - you need a function to have a domain.


Hope that helps.

7 0
4 years ago
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