Answer:
c = 2 , d = 3
Step-by-step explanation:
(2y^3)(9y^3) = 18y^6
First we'll do two basic steps. Step 1 is to subtract 18 from both sides. After that, divide both sides by 2 to get x^2 all by itself. Let's do those two steps now
2x^2+18 = 10
2x^2+18-18 = 10-18 <<--- step 1
2x^2 = -8
(2x^2)/2 = -8/2 <<--- step 2
x^2 = -4
At this point, it should be fairly clear there are no solutions. How can we tell? By remembering that x^2 is never negative as long as x is real.
Using the rule that negative times negative is a positive value, it is impossible to square a real numbered value and get a negative result.
For example
2^2 = 2*2 = 4
8^2 = 8*8 = 64
(-10)^2 = (-10)*(-10) = 100
(-14)^2 = (-14)*(-14) = 196
No matter what value we pick, the result is positive. The only exception is that 0^2 = 0 is neither positive nor negative.
So x^2 = -4 has no real solutions. Taking the square root of both sides leads to
x^2 = -4
sqrt(x^2) = sqrt(-4)
|x| = sqrt(4)*sqrt(-1)
|x| = 2*i
x = 2i or x = -2i
which are complex non-real values
Answer:
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Step-by-step explanation:
-2x=5 is the right answer
Answer: 1≤x<12
Step-by-step explanation:
The values that are greater than or equal 1 but less than 12 are values ranging from 1 to 11 i.e {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}. Note that 1 is included because the number is greater than or equal to 1 (1 included) and less than 12(12 not included).
x≥1 and x< 12
if x≥1, it means 1≤x
combining both inequality
1≤x and x<12
1≤x<12
The interval notation that represent the set S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11} will be given as S = {x: 1≤x<12}