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 .
Answer:
5 terms
Step-by-step explanation:
nth term of the sequence =n^2 + 20
an= n^2 + 20
1st term when n= 1
1^2 + 20= 20
2nd term n= 2
2^2 + 20=24
3rd term when n= 3
3^2 + 20= 29
4th term when n= 4
4^2 + 20= 36
5th term when n= 5
5^2 + 20 =45
6th term when n= 6
6^2 + 20=56
Hence, terms in the sequence are less than 50 are first 5 terms
Answer:
y = 5.595090517 or approximately 5.6
Step-by-step explanation:
Tan47 = 6/y
y x Tan47/Tan47 = 6/Tan47
y = 5.595090517 or approximately 5.6
X<9
The answer is x<9 because you first subtract 4 from both sides then divide both sides by -3. Which then would give you 9.
Hello!
It is given information that this triangle is a right triangle, meaning that the Pythagorean Theorem can apply here.
We can set the length of a leg as an unknown variable 
.
Remember that it is the whole 
term being squared, meaning that it'll simplify to 
, not 
.
This means we can plug back in to find the sides.
The hypotenuse would be 
, or 4.2.
And the legs would be 1.4 and 4, respectively.
Hope this helps!
