Answer:
where is the problem
Step-by-step explanation:
Step-by-step explanation:
Please gimme brainliest
I hope it's correct
<u>Given</u><u> </u><u>that</u><u>:</u><u>-</u>
→ 1 sandwich = 2 slices of bread.
→ 1 hiker = 1 sandwich.
→ Then we have to find number of bread slices for n hikers .
→ Number of bread slices for 1 hiker = 2
→ Number of bread slices for 2 hikers = 2 × 2
→ For 3 hikers = 3 × 2
So in similar way
→ Number of bread slices for n hikers = 2×n → 2n
So <u>2</u><u>n</u><u> </u> is the answer.
Answer:
second table
Step-by-step explanation:
Out of the 8 options on the spinner, 2 of them are 0's, 1 of them is a 1, 2 of them are 2's and 3 of them are 3's so the probability of spinning a 0, 1, 2 or 3 is 2/8, 1/8, 2/8 or 3/8 which becomes 0.25, 0.125, 0.25 or 0.375 respectively. Therefore, the answer is the second table.
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.
