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.
Yess, such fun so
q=number of quarter
n=number of nickles
q+n=63
9.15=915 cents
25 cents=1 q
5 cents=1 n so
915=25q+5n
915=5(5q+n)
divide by 5
183=5q+n
we also have q+n=63
subtract q from both sides
n=63-q
subsitute 63-q for n in second equation
183=5q+63-q
add like terms
183=4q+63
subtract 63 from both sides
120=4q
divdie by 4
30=q
there were 30 quarters
subsitute
63=30+n
subtract 30
33=n
30 quarters
33 nicles
Answer:
A.(-2, 0)
C. (-1.4)
Step-by-step explanation:
we know that
If a point lie on the line, then the point must satisfy the equation of the line (makes the equation true)
we have
subtract 7 both sides
divide by 2 both sides
Substitute the value of x and the value of y of each point in the linear equation and analyze the result
<u><em>Verify each point</em></u>
case A) we have
(-2, 0)
For x=-2, y=0
substitute
---> is true
so
the point lie on the line
case B) we have
(1, 3)
For x=1, y=3
substitute
---> is not true
so
the point not lie on the line
case C) we have
(-1, 4)
For x=-1, y=4
substitute
---> is true
so
the point lie on the line
case D) we have
(1, -4)
For x=1, y=-4
substitute
---> is not true
so
the point not lie on the line
case E) we have
(0, -1)
For x=0, y=-1
substitute
---> is not true
so
the point not lie on the line
10 is the answer. Hope that helped
The solution to the problem is as follows:
cot θ = 1 / tanθ
<span>cot θ tanθ = tanθ / tanθ = 1
</span>
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