The answer would be 1.40. The 3 significant figure are 1.39 and it wants us to round so we look at the 8 in 1.3981 and since 8 > 5 we know to round up.
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:
n = -4
Step-by-step explanation:
You must write and solve a symbolic equation.
n - 6 = 5(n + 2), or
n - 6 = 5n + 10
Combining like terms, we get:
-16 = 4n,
and so n = -4
Answer:
Step-by-step explanation:
To prove the given identity, we solve the left hand side and right hand side expressions and show that they are equal.
So we get
