5/6 × 6 = 5 so 5/6 times 12= 10
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:
no solution.
Step-by-step explanation:
Answer:
x=-5/11, y=-9/11. (-5/11, -9/11).
Step-by-step explanation:
3x+2y=-3
y=4x+1
---------------
3x+2(4x+1)=-3
3x+8x+2=-3
11x=-3-2
11x=-5
x=-5/11
y=4(-5/11)+1
y=-20/11+11/11
y=-9/11
Answer:
Answer is 12
Step-by-step explanation:
f(g(-2))
=f((-2)^2+2)
=f(6)
=-1/2×6^2+5×6
=12