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Wittaler [7]
3 years ago
7

Determine whether each ordered pair is a solution to the inequality x+y>−2.

Mathematics
1 answer:
sweet [91]3 years ago
4 0
<h3>3 Answers:</h3>
  • Choice A.  (10, -1)
  • Choice B.  (-8, 9)
  • Choice D.  (6, -3)

===================================

Explanation:

If we plug the coordinates of point A into the inequality, then we get

x+y > -2

10 + (-1) > -2

9 > -2

That last inequality is a true statement since 9 is to the right of -2 on the number line. That means (10,-1) is a solution. Choice A is one of the answers

Choices B and D are also answers for similar reasons.

Something like choice C is not a solution because

x+y > -2

-1+(-9) > -2

-10 > -2

which is false

You should find that choice E is false as well.

---------------------

If you graphed the inequality and all of the points mentioned (see below), then you can visually confirm the answers. Notice how points A, B and D are in the blue shaded region which is the solution set.

The point E on the boundary does not count as a solution. This is due to the lack of "or equal to" portion of the inequality sign. That visually shows point E is not a solution. Point C isn't a solution either as it's nowhere near the blue shaded region.

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Step-by-step explanation:

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Combine Like Terms:<br> 10z + 2z - 2
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Step-by-step explanation:

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A cone is 10 inches tall and has a radius of 3 inches. What is the cone's volume?
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In each problem 7 through 14, verify that each given function is a solution of the differential equation. Y" - y = 0; y1(t) = et
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Answer:

Since L.H.S = R.H.S = 0, for both y_{1} (t) = e^{t} and y_{2} (t) = cosh(t), y₁ and y₂ both satisfy the equation y" - y = 0 and are thus solutions to the differential equation.

Step-by-step explanation:

To check whether the given functions are solutions the given differential equation, we differentiate the functions and then insert it into the given equation.

So y" - y = 0 and  

y_{1} (t) = e^{t}\\y_{1}' (t) = e^{t}\\ y_{1}" (t) = e^{t}

Substituting these values of y and y" into the left hand side of the equation, we have

y" - y  

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Since L.H.S = R.H.S

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When

y_{2} (t) = cosh(t)\\ y_{2}'(t) = sinh(t) \\y_{2}"(t) = cosh(t)

Substituting y and y" into the left hand side of the equation, we have

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Since L.H.S = R.H.S

So, y_{2} (t) = cosh(t) is a solution of the differential equation.

4 0
4 years ago
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