The number of ways for which she could pick four colours if green must be one of them is; 10 ways.
<h3>How many ways can she picks four colours if green must be there?</h3>
It follows from the task that there are 6 colours in total that she could pick from.
Hence, since she needs four colours with green being one of them, it follows that she only has 3 colours to pick from 5.
Hence, the numbers of possible combinations is; 5C3 = 10 ways.
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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.
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Answer:
all are functions except Relation 2
Step-by-step explanation:
If any input value (first of an ordered pair) is used more than once, the relation is not a function. Output values can be used as often as you like.
Only Relation 2 is not a function.
