Because 5/8 + 6/10 is 49/40 and 49/40 is equal to 1 9/40
No, a cubic equation can not have three complex roots. This is because it turns twice and one end goes to positive infinity and one end goes to negative infinity. Thus, one of these MUST cross the x-axis at some point, meaning y = 0 and a real root exists.
Yes, a cubic equation can have three real roots if it cuts the x-axis three times.
-2x - 13 = -3x - 5
Step 1: Combine like terms
x's go with x's (-2x and -3x). To do this add 2x to both sides
(-2x + 2x) - 13 = (-3x + 2x) - 5
(0) - 13 = (-x) - 5
- 13 = -x - 5
normal numbers go with normal numbers ( -13 and -5). To do this add 5 to both sides
(- 13 + 5) = -x + (- 5 + 5)
-8 = -x + (0)
-8 = -x
Step 2: Isolate x by dividing -1 to both sides (this will take the negative sign away from the x)

8 = x
x= 8
Hope this helped!
Answer:
Where's the image?
Step-by-step explanation: