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.
Answer:
b(-10) = 6
Step-by-step explanation:
Step 1: Define
b(x) = |x + 4|
b(-10) is x = -10
Step 2: Substitute and Evaluate
b(-10) = |-10 + 4|
b(-10) = |-6|
b(-10) = 6
Answer: 20
Step-by-step explanation: Lets break down the problem;
You always use multiplication first when solving an expression.
26 - 1 x 6
1 x 6 = 6
26 - 6 = 20
Remark
You have to read this carefully to know what way to write it.
One way you could write it would be
20 * x = 18 where x is the fraction. Now all you need do is divide.
Divide both sides by 20
20/20 * x = 18/20

But you should reduce this fraction. The way to do that is to factor 18 and 20 into primes.
18: 2 * 3 * 3
20: 2 * 2 *5
Cancel out the Highest common factor or 2.
You are left with
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