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: For a is 19/15 or 1 4/15
for b I have no clue
Step-by-step explanation: add 3/5 and 2/3
you will get 19/15. In mixed fraction is 1 4/15
B. 45 items x $2 = 90$. 90 + the 25 from the beginning = 115
The change in temperature is -28°F
So, Option D is correct.
Step-by-step explanation:
Temperature at day time = 72°F
Temperature at night = 44°F
What number represents the change in number?
Change in temperature = Temperature at night - Temperature at Day
Change in temperature = 44°F-75°F
Change in temperature = -28°F
The change in temperature is -28°F
So, Option D is correct.
Keywords: Solving Expressions
Learn more about Solving Expressions at:
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Answer:
63
Step-by-step explanation:
She finds her average resting heart rate to be 62 after 5 days.
That means, average of the 5 heart rate pulse=62
If the last day was 58 and all the other days were the same.
Let x be her pulse on the other days,
Then,
Her pulse on all the other days is 63.