Answer:
i think is the last one
Step-by-step explanation:
The quadratic equations and their solutions are;
9 ± √33 /4 = 2x² - 9x + 6.
4 ± √6 /2 = 2x² - 8x + 5.
9 ± √89 /4 = 2x² - 9x - 1.
4 ± √22 /2 = 2x² - 8x - 3.
Explanation:
- Any quadratic equation of the form, ax² + bx + c = 0 can be solved using the formula x = -b ± √b² - 4ac / 2a. Here a, b, and c are the coefficients of the x², x, and the numeric term respectively.
- We have to solve all of the five equations to be able to match the equations with their solutions.
- 2x² - 8x + 5, here a = 2, b = -8, c = 5. x = -b ± √b² - 4ac / 2a = -(-8) ± √(-8)² - 4(2)(5) / 2(2) = 8 ± √64 - 40/4. 24 can also be written as 4 × 6 and √4 = 2. So x = 8 ± 2√6 / 2×2= 4±√6/2.
- 2x² - 10x + 3, here a = 2, b = -10, c = 3. x =-b ± √b² - 4ac / 2a =-(-10) ± √(-10)² - 4(2)(3) / 2(4) = 10 ± √100 + 24/4. 124 can also be written as 4 × 31 and √4 = 2. So x = 10 ± 2√31 / 2×2 = 5 ± √31 /2.
- 2x² - 8x - 3, here a = 2, b = -8, c = -3. x = -b ± √b² - 4ac / 2a = -(-8) ± √(-8)² - 4(2)(-3) / 2(2) = 8 ± √64 + 24/4. 88 can also be written as 4 × 22 and √4 = 2. So x = 8 ± 2√22 / 2×2 = 4± √22/2.
- 2x² - 9x - 1, here a = 2, b = -9, c = -1. x = -b ± √b² - 4ac / 2a = -(-9) ± √(-9)² - 4(2)(-1) / 2(2) = 9 ± √81 + 8/4. x = 9 ± √89 / 4.
- 2x² - 9x + 6, here a = 2, b = -9, c = 6. x = -b ± √b² - 4ac / 2a = -(-9) ± √(-9)² - 4(2)(6) / 2(2) = 9 ± √81 - 48/4. x = 9 ± √33 / 4 .
What is the third term of the sequence defined by the recursive rule f(1)=3 f(n)=f(n-1)+4
Need f(2):
f(2)=f(2-1)+4
f(2)=f(1)+4
f(2)=(3)+4=7
FIND f(3):
f(3)=f(3-1)+4
f(3)=f(2)+4
f(3)=(7)+4
f(3)=11
<span>cy+3=6d-2y
cy + 2y = 6d - 3
(c + 2)y = 6d - 3
y = (6d - 3)/(c + 2)</span>
