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.
Let x = number of months.
Let y = total number of fish in pond.
6% of 3,000 = 180 fish
We can use an equation to help us with this problem.
y = 3,000 + 180x - 400x. Combine like terms.
y = 3,000 - 220x. Plug 9 in for x.
y = 3,000 - 220(9). Solve for y
y = 1,020.
There are 1,020 fish in the pond after 9 months.
The answer is 692.857143
Hope this helps! :)
Answer:
There are 5 black counters in the bag.
Step-by-step explanation:
15 green counters in the bag
The proportion of green counters is given by:
So, we have that, the total is x. So
There are 30 total counters.
How many black counters are in the bag ?
A sixth of the counters are black. So
There are 5 black counters in the bag.
Answer:
2x-11
Step-by-step explanation:
The product of something means multiplication, and since we are only given one number we can use the variable x to represent it. And minus 11 is obviously subtraction, so we can put together the expression 2x-11.
