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:
account settings
Step-by-step explanation:
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To find the area of a prism, you calculate the area of the cross-section and multiply that number by the length. The cross-section of this prism is a trapezium, and the formula for this is
.
Insert your given numbers to find out the area of this prism's cross-section:
Now, multiply 90 by the length, 10cm, to find out the volume of the prism;
90×10=900cm³
Therefore, your answer id the third option.
Answer:
86
Step-by-step explanation:
the rectangle portion is 18*4= 72
1 triangle portion is 1/2bh where the base is 4 and height is 3.5
1/2*14=7
the 2nd triangle as the same proportions, so it is also 7
72+7+7=86
We have the following equation:
2x2-3x + 1
Using resolver we have:
x = (- b +/- root (b2-4ac)) / (2a)
Substituting values we have:
x = (- (- 3) +/- root ((- 3) 2-4 (2) (1))) / (2 (2))
Rewriting:
x = (3 +/- root (9-8)) / (4)
x = (3 +/- root (1)) / (4)
b2-4ac = 1> 0
root(b2-4ac) = 1 < 3
Answer:
Two positive real solutions
option C
