The answer is 70 the third option.
Answer: You can do this!
Step-by-step explanation: For D. The x= -1 and the y= 1 for E same thing the x= 7 and the y= 1 so look for the 7 on the x-axes and the one on the y-axes get it. I believe that you got this!
Hope this helped love<3
Answer:
83
supplementary is two angles that add up to 180 so 180 - 97= 83
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.
<span>We simplify the equation to the form, which is simple to understand
<span>4x+8-2x=20
We move all terms containing x to the left and all other terms to the right.
<span>+4x-2x=+20-8
We simplify left and right side of the equation.
<span>+2x=+12
We divide both sides of the equation by 2 to get x.
<span>x=6</span></span></span></span></span>
