She added the check fee instead of subtracting it
Answer:
Roots are not real
Step-by-step explanation:
To prove : The roots of x^2 +(1-k)x+k-3=0x
2
+(1−k)x+k−3=0 are real for all real values of k ?
Solution :
The roots are real when discriminant is greater than equal to zero.
i.e. b^2-4ac\geq 0b
2
−4ac≥0
The quadratic equation x^2 +(1-k)x+k-3=0x
2
+(1−k)x+k−3=0
Here, a=1, b=1-k and c=k-3
Substitute the values,
We find the discriminant,
D=(1-k)^2-4(1)(k-3)D=(1−k)
2
−4(1)(k−3)
D=1+k^2-2k-4k+12D=1+k
2
−2k−4k+12
D=k^2-6k+13D=k
2
−6k+13
D=(k-(3+2i))(k+(3+2i))D=(k−(3+2i))(k+(3+2i))
For roots to be real, D ≥ 0
But the roots are imaginary therefore the roots of the given equation are not real for any value of k.
Answer:
The required function is:
Step-by-step explanation:
We have to represent the given scenario as an equation or function
Let x be the number of miles driven in a week
Let C(x) be the function of the number of miles driven
As it is given that charges are 150 per week, these charges are constant so they will be used as it is.
It is also given that the cost of car is 0.45 per mile so for x miles the cost will be:
0.45x
Combining both terms, we get
Hence,
The required function is:
Answer:
B. x = -8
Step-by-step explanation:
-4(2x + 3) = 2x + 6 - (8x + 2)
-8x -12 = 2x + 6 - 8x -2
(now, "-8" in both terms is cancelled):
-12 = 2x + 6 - 2
(leave 2x alone in second term):
-12 -6 +2 = 2x
-16 = 2x
-16/2 = x
-8 = x
