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>Given that B=(1,2,3,4) how many subsets have exactly two elements?
4C2 =
4! 4! 4 x 3
------------ = -------------- = --------------- = 6
2! (4-2)! 2! 2! 2 x 1
answer is 6</span>
Answer:
Area of triangle: 18
Area of semicircle: (9pi)/2
perimeter of triangle: 20.49
perimeter of semicircle: 3pi+6
perimeter of whole thing: 23.91/ 6+8.49+3pi
Step-by-step explanation:
triangle area: 1/2bh, 1/2*6*6=18
Area of circle: pir^2, 3^2*pi, 9pi, semicircle means half, 9pi/2
There would be infinite solutions if no initial-value problem is specified. For example, take the differential equation written below:
dy = 3dx
When you differentiate that, the equation would become:
y - y₀ = 3(x - x₀)
Now, there can be arbitrary values of x₀ and y₀. If no initial-value problem is specified, you cannot solve the problem because there are infinite solution. Example of an initial value problem is: when x₀ = 2, y₀ = 2. If we had that, we can find a solution to the differential equation.
Answer:
12
Step-by-step explanation:
4 x 3 = 12
