Step-by-step explanation:
For quadratic equation ax^2 + bx + c = 0 to have two distinct real roots,
b^2 - 4ac must be positive.
b^2 - 4ac > 0
(k - 3)^2 - 4(3 - 2k) > 0
k^2 - 6k + 9 - 12 + 8k > 0
k^2 + 2k - 3 > 0
This is all percentages so...
4/9 = 44%
2/7 = 29%
1/10 = 10%
Total = 83%
So 17% of 630 is left over which = 107
Answer:
<u>Given rhombus ABCD with</u>
- m∠EAD = 67°, CE = 5, DE = 12
<u>Properties of a rhombus:</u>
- All sides are congruent
- Diagonals are perpendicular
- Diagonals are angle bisectors
- Diagonals bisect each other
<u>Solution, considering the above properties</u>
- 1. m∠AED = 90°, as angle between diagonals
- 2. m∠ADE = 90° - 67° = 23° as complementary of ∠EAD
- 3. m∠BAE = 67°, as ∠BAE ≅ ∠EAD
- 4. AE = CE = 5, as E is midpoint of AC
- 5. BE = DE = 12, as E is midpoint of BD
Answer:
p = 9
Step-by-step explanation:
Answer:
it 12 inches u can thank me later
