K² + 13k + 30 = 0
This is a quadratic equation. Let's attempt solving it by factorization.
Coefficient of k² = 1, last term = 30
Multiply these two together 1*30 = 30
We think of two numbers that multiply to give 30 and add to give the middle term 13.
The two numbers are 3 and 10.
3*10 = 30, 3 + 10 = 13.
we replace the 13k with 3k + 10k
k² + 13k + 30 = 0
k² + 3k + 10k + 30 = 0
k(k + 3) + 10(k + 3) = 0
(k + 3)(k + 10) = 0
(k + 3) = 0 or (k + 10) = 0
k = 0 - 3 or k = 0 - 10
k = -3, or -10
I hope this helps.
Answer: x = (1,9]
Step-by-step explanation: If x is all real numbers greater than 1 (simply: x is greater but not equal to 1) then it is 1 < x. And if x is less than or equal to 9 then its x ≤ 9. To translate 1 < x ≤ 9 into interval notation, you need to keep in mind that parentheses are for these symbols: <>. And brackets are for these symbols: ≤≥. With that in mind, your answer is (1,9]. Also, keep in mind that interval notation goes from lowest value number to highest, so you wouldn’t write this notation as [9,1).
Answer:
Step-by-step explanation:
1. Three zeroes equal 1,000
2. One zero equal 10
3. One zero equal 10
4. Look at the value without the zeroes
5. Three zeroes equal 1,000
6. Look at the value without the zeroes
7. Since 890 has one zero, one more zero equal 10
8. Look at the value without the zeroes
9. Same as 8.
10. Three zeroes equal 1,000
11. Look at the value without the zeroes
12. Three zeroes equal 1,000
13. Same as 11.
14. Same as 13.
15. N/A
16. I am not sure about this one cause I don't know if there are more zeroes or not, but same as 13.
I’m going to say pre-algebra because that’s when I learned it in my school system but you may also learn it in algebra as well, College ready math can start in 8th grade math but I would think it would be in highschool.
