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Grace [21]
2 years ago
9

How many integers in the set of all integers from 1 to 200, inclusive, are not the cube of an integer

Mathematics
1 answer:
ss7ja [257]2 years ago
4 0

Cubing of integers from 1 to 5 will have their cubes in the interval 1 to 200. There are such 5 positive numbers (1,2,3,4,5).

To calculate the variety of integers, locate and subtract the integers of the hobby after which subtract 1. As evidence of the concept, calculate the wide variety of integers that fall between five and 10 on more than a few lines. We realize there are four (6, 7, 8, 9).

All entire numbers are integers (and all herbal numbers are integers), however now not all integers are whole numbers or natural numbers. as an example, -five is an integer but not an entire variety or a natural quantity. An integer is an entire variety (not a fractional number) that can be fine, bad, or 0. Examples of integers are: -5, 1, 5, 8, 97, and 3,043. Examples of numbers that are not integers are: -1.forty three, 1 3/four, three.14, . 09, and five,643.1.

Learn more about integers here: brainly.com/question/17695139

#SPJ4

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A

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Step-by-step explanation:

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3 years ago
What is the solution to the following system?<br> X+2y+z=9<br> X-y+3z=13<br> 2z=10
Vinvika [58]

Answer:  x = 0; y = 2; z = 5

Step-by-step explanation:

x+2y+z=9

x-y+3z=13

2z=10

==

Start with the last equation:  2z = 10, therefore z = 5 [easy]

Let's rewrite the second equation with 1) z=5, and 2) rearrange to find x:

x-y+3z = 13

x - y +3*5 = 13

x = y-2

Now use the values for x [(y-2)] and y [5] in the first equation:

X+2y+z=9

(y-2) +2y + 5 = 9

3y +3 = 9

3y = 6

y = 2

Now that we have y [2] and z [5], let's use the first equation to solve for x:

x+2y+z=9

x + 4 + 5 = 9

x + 9 = 9

x = 0

====

Try the three values (x=0, y=2, and z=5) in any of the formulas.  The result should/will equal the number shown.

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What must be true for line EF to not intersect plane L?
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Answer:

If line EF is parallel to line CD, it will not intersect plane L.

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