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.
Answer
8(40-y) + 6y =300
320 - 8y + 6y =300
-2y=-20
y=10
This leaves x to be 30.
You should spend 30 minutes on machine 1 and 10 minutes on machine 2.
Step-by-step explanation:
<h3>Answer:</h3>
$808.38
<h3>Explanation:</h3>
The formula for the payment amount (A) on principal P at interest rate r compounded monthly for a loan period of t years is ...
... A = P(r/12)/(1 -(1 +r/12)^(-12t))
For the main loan, the payment is ...
... A = 0.80·145000·(.0475/12)/(1 -(1 +.0475/12)^(-12·30)) = 605.11
For the piggyback loan, the payment is ...
... A = 0.20·145000·(.07525/12)/(1 -(1 +.07525/12)^(-12·30)) = 203.27
So, the total of monthly payments for the two loans is ...
... $605.11 +203.27 = $808.38
Im not sure but i think its 2/3
hope this helps
