18<x is the answer you are looking for. Glad to help.
Hey there! I'm happy to help!
To find the surface area of a sphere, here is what you do.
You square the radius.
4²=16
You multiply by 4.
16×4=64
And you multiply by pi!
64×π=64π
Therefore, the surface area of the sphere is A. 64π units². It's that easy!
Now you can find the surface area of a sphere! Have a wonderful day! :D
Answer:
2x-y= -3
Step-by-step explanation:
2x=y-3
2x-y= -3
There is an app called photomath and u can take pictures of math problems and get the answer
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.
