Answer:
C. y= 6.18x + 10.9
Step-by-step explanation:
Every game that he plays (x) costs 6.18, for example, if he plays 3 games it would be 6.18 times 3 (games)
Now you have to add your admission, 10.9
To enter he needs to pay the admission ( 10.9 ) and also each game ( x ) he plays ( 6.18 )
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:
assuming the squares are 1cm by 1cm
area 7
perimeter 16
Step-by-step explanation:
it just works
First substitute, then distribute, then add, then simply
