For a quadratic of the form
![f(x)=ax^2+bx+c](https://tex.z-dn.net/?f=f%28x%29%3Dax%5E2%2Bbx%2Bc)
, we have the quadratic formula
![x=\dfrac{-b \pm \sqrt{b^2 -4ac} }{2a}](https://tex.z-dn.net/?f=x%3D%5Cdfrac%7B-b%20%5Cpm%20%5Csqrt%7Bb%5E2%20-4ac%7D%20%20%7D%7B2a%7D)
,
where a is the coefficient (number before the variable) of the squared term, b is the coefficient of the linear term, and c is the constant term.
So, given
![2x^2-16x+27=0](https://tex.z-dn.net/?f=2x%5E2-16x%2B27%3D0)
, we can get that
![a=2, \ b=-16](https://tex.z-dn.net/?f=a%3D2%2C%20%5C%20b%3D-16)
, and
![c=27](https://tex.z-dn.net/?f=c%3D27)
. We substitute these numbers into the quadratic formula above.
![x=\dfrac{-(-16) \pm \sqrt{(-16)^2 -4(2)(27)} }{2(2)}](https://tex.z-dn.net/?f=x%3D%5Cdfrac%7B-%28-16%29%20%5Cpm%20%5Csqrt%7B%28-16%29%5E2%20-4%282%29%2827%29%7D%20%7D%7B2%282%29%7D%20)
![x=\dfrac{16 \pm \sqrt{(256 -216)} }{4}](https://tex.z-dn.net/?f=x%3D%5Cdfrac%7B16%20%5Cpm%20%5Csqrt%7B%28256%20-216%29%7D%20%7D%7B4%7D%20)
![x=\dfrac{16 \pm \sqrt{40} }{4}](https://tex.z-dn.net/?f=x%3D%5Cdfrac%7B16%20%5Cpm%20%5Csqrt%7B40%7D%20%7D%7B4%7D%20)
![x=\dfrac{16 \pm 2\sqrt{10} }{4}](https://tex.z-dn.net/?f=x%3D%5Cdfrac%7B16%20%5Cpm%202%5Csqrt%7B10%7D%20%7D%7B4%7D%20)
![x=4+ \frac{\sqrt{10}}{2}, \ x=4- \frac{\sqrt{10}}{2}](https://tex.z-dn.net/?f=x%3D4%2B%20%5Cfrac%7B%5Csqrt%7B10%7D%7D%7B2%7D%2C%20%5C%20x%3D4-%20%5Cfrac%7B%5Csqrt%7B10%7D%7D%7B2%7D)
This is our final answer.
If you've never seen the quadratic formula, you can derive it by completing the square for the general form of a quadratic. Note that the
![\pm](https://tex.z-dn.net/?f=%5Cpm)
symbol (read: plus or minus) represents the two possible distinct solutions, except for zero under the radical, which gives only one solution.
Answer:
See explanation
Step-by-step explanation:
1. BK is an angle B bisector, then
(definition of angle bisector)
2. BM = MK, then
triangle BMK is isosceles triangle with base BK.
3. Angles adjacent to the base of isosceles triangle are congruent, then
![\angle MBK \cong \angle BKM](https://tex.z-dn.net/?f=%5Cangle%20MBK%20%5Ccong%20%5Cangle%20BKM)
Note that angle MBK is the same as angle CBK.
4. By substitution property,
![\angle ABK \cong \angle BKM](https://tex.z-dn.net/?f=%5Cangle%20ABK%20%5Ccong%20%5Cangle%20BKM)
5. By alternate interior angles theorem,
if
, then ![AB\parallel KM](https://tex.z-dn.net/?f=AB%5Cparallel%20KM)
Answer:
A
Step-by-step explanation:
Question:
Which of the following expressions represents the distance between 1.7 and −1/2 on a number line?
Choose 1 answer:
![A.\ |1.7 - 1/2|](https://tex.z-dn.net/?f=A.%5C%20%7C1.7%20-%201%2F2%7C)
![B.\ |1.7 -(-1/2)|](https://tex.z-dn.net/?f=B.%5C%20%7C1.7%20-%28-1%2F2%29%7C)
C. None of the above
Answer:
![B.\ |1.7 -(-1/2)|](https://tex.z-dn.net/?f=B.%5C%20%7C1.7%20-%28-1%2F2%29%7C)
Step-by-step explanation:
Given
Numbers: 1.7 and -1/2
Required
Determine the distance
To calculate the distance, we start by calculating the difference between the given numbers;
![Difference = 1.7 - (-1/2)](https://tex.z-dn.net/?f=Difference%20%3D%201.7%20-%20%28-1%2F2%29)
Then, we calculate the distance;
<em>In this case, the distance is meant to be positive; So, we'll introduce the symbol for absolute value;</em>
Hence,
![Distance = |Difference|](https://tex.z-dn.net/?f=Distance%20%3D%20%7CDifference%7C)
Substitute
for Difference;
![Distance = |1.7 - (-1/2)|](https://tex.z-dn.net/?f=Distance%20%3D%20%7C1.7%20-%20%28-1%2F2%29%7C)
<em>Hence, the correct option is B</em>