It could sometimes be proportional. The reasoning here is that if all the items have different prices it won't be, however if the items you are ordering all have the same price they will be!
Answer:
we conclude that the 'bait and switch' technique is not a promotional tactic used by a seller.
Hence, 'a' is the correct option.
Step-by-step explanation:
From the given options, the 'bait and switch' technique is not a promotional tactic used by a seller.
'Bait and switch' is a deceptive sale practice using which the sells try to attract (bait) the customers by offering attractive prices on certain items, but when the customers tend to go to the shop to buy the items, they witness the unavailability of the goods, or find the prices go higher compared to what they had been offered.
Now, since the customers are already present at the shop, the sellers try to pressurize the customer so that they could buy something else.
Therefore, we conclude that the 'bait and switch' technique is not a promotional tactic used by a seller.
Hence, 'a' is the correct option.
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.
