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Alex17521 [72]
3 years ago
8

Use the x-intercept method to find all real solutions of the equation -9^3-72^2-96x+36=3x^3+x^2-3x+8

Mathematics
1 answer:
Vadim26 [7]3 years ago
7 0

Answer:

 x ≈ 0.290640965127

Step-by-step explanation:

y subtracting the right side of the equation, we get a function that is zero at a real solution for x. The only x-intercept is at approximately 0.290640965127.

___

The graph shows the x-intercept to be 0.2906. The value above was obtained by Newton's method iteration. The roots of this cubic will all be irrational.

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34 = 2 ( x + 2 + 3x - 4 )

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34 = 8x - 4

38 = 8 x

38/ 8 = x

thus , x = 4.75 inches

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Michael estimated his mass as 8 kilograms.Is his estimate reasonable?Justify your answer.
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Esteban has a big jar of change in his room. He has 600 coins total, and 240
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Question 5 and 6 please help me
sp2606 [1]

Problem 5

The function is continuous for the given domain x \ge 6

This is because y = (-5/6)x+5 is itself continuous, and any interval subset of this function is also continuous. We can plug in any real number that is equal to 6 or larger, and get some y output. If we plugged in x = 6, then we'd get

y = (-5/6)x+5

y = (-5/6)*6 + 5

y = -5+5

y = 0

This is the largest y value possible. Why? Because y = (-5/6)x+5 has a negative slope, so the graph is going downhill as you read it from left to right. As x gets bigger, y gets smaller. The smallest x value allowed in the domain produces the largest y value in the range. There is no smallest y value as the y values keep going down forever.

The range is therefore y \le 0

In interval notation, you can write the range as (-\infty, 0]. The square bracket indicates "include this endpoint as part of the range".

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

Problem 6

The function is discrete for this given domain. The domain itself is a discrete list of values. We cannot plug in values between say 0 and 2. We can only substitute one of those values from the list given. Consequently, the y values will also be a list, and not an interval like problem 5 had.

-----------

If you plugged in x = -4, then you should get...

y = (-1/2)*(-4)+2

y = 2+2

y = 4

So the input x = -4 lead the output y = 4

Repeat for x = -2

y = (-1/2)x+2

y = (-1/2)*(-2)+2

y = 1+2

y = 3

and the same for x = 0 as well

y = (-1/2)x+2

y = (-1/2)*0 + 2

y = 0 + 2

y = 2

and x = 2 also

y = (-1/2)x+2

y = (-1/2)*2 + 2

y = -1+2

y = 1

Finally, plug in x = 4

y = (-1/2)x+2

y = (-1/2)*4+2

y = -2+2

y = 0

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

If we plugged each of these x values {-4, -2, 0, 2, 4} one at a time into the equation y = (-1/2)x+2, then we get this list of values {4, 3, 2, 1, 0}

Sorting the values from smallest to largest, we have this range {0, 1, 2, 3, 4}

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