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:
Let 'x' and 'y' be two different numbers.
Leila says that 75% of a number will always be greater than 50% of a number. The inequality that represents this statement is the following:
0.75x > 0.5y
Let x = 100 and y=200. We have that:
0.75(100) > 0.5(200)
75 > 100 ❌ INCORRECT ❌
Given that we found a case in which 75% of a number is not greater than 50% of a number, we can conclude that Leila's claim is incorrect.
answer
Step-by-step explanation:
Answer:
<em>A) 84x-16y-10.6</em>
2(14x-3y+28x+12.2-5y-17.5)
2(48x-3y+12.2-5y-17.5)
2(48x-8y+12.2-17.5)
2(48x-8y-5.3)
<em>84x-16y-10.6</em>
15% = 0.15
0.15 • 50 = 7.5
50 + 7.5 = 57.5
