Answer:
x = 5, y = -2
Step-by-step explanation:
<u>Given equations:</u>
<u>From the first equation we get:</u>
<u>Substitute x in the second equation, find y:</u>
- 3x - y = 17
- 3*(1-2y) - y = 17
- 3 - 6y - y = 17
- -7y = 17 - 3
- -7y = 14
- y = -14/7
- y = -2
<u>Find x:</u>
- x = 1 - 2y
- x = 1 - 2*(-2)
- x = 1 + 4
- x = 5
90 round to the nearest hundredth
Did u type the equation wrong none of these graphs match I believe the intercept would be 2
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:
no
Step-by-step explanation:
