Answer:
Infinite pairs of numbers
1 and -1
8 and -8
Step-by-step explanation:
Let x³ and y³ be any two real numbers. If the sum of their cube roots is zero, then the following must be true:
![\sqrt[3]{x^3}+ \sqrt[3]{y^3}=0\\ \sqrt[3]{x^3}=- \sqrt[3]{y^3}\\x=-y](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bx%5E3%7D%2B%20%5Csqrt%5B3%5D%7By%5E3%7D%3D0%5C%5C%20%5Csqrt%5B3%5D%7Bx%5E3%7D%3D-%20%5Csqrt%5B3%5D%7By%5E3%7D%5C%5Cx%3D-y)
Therefore, any pair of numbers with same absolute value but different signs fit the description, which means that there are infinite pairs of possible numbers.
Examples: 1 and -1; 8 and -8; 27 and -27.
Why this is so easy
Ok so, 75/5=15
15*7 =105 so the answer is 105 pages
Answer:
Step-by-step explanation:
The wording on this is not the best. It sounds like the 1 zero has even multiplicity (that's because of where the modifier is). On top of that it has an odd power. You could try this. y =x*(x^2+1)^2
The problem is not with the power. It gives x^5. The problem is with the multiplicity of the one place where it crosses. (X^2 + 1) does factor, but it gives a complex root. I'm not sure that's allowed. However, it is the best I can do.
<h3>
Step-by-step explanation:</h3>
Quadrilateral are four sided shapes. e.g
There are 2 sides in a quadrilateral and each of them are 180° shown above. Angles in a quadrilateral add up to 360°.
For example: The example is shown in the picture above.
Answer:
I think I just posted that last time. Check my timeline
