Answer:
A.(26)X =1
26x =1
divide both sides by 26 to make x the subject
26x/1=1/26
x = 1/26
Step-by-step explanation:
B 50(x) =1
50x =1
x =1/50
(-3) + (-5) + 9
(-3) + (-5) = -8
-8 + 9
= 1
Answer:
704
Step-by-step explanation:
8 x 8 x 8 I think its the answer hope it helps
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:
decimals
Step-by-step explanation:
