Answer:
Shrink:
a=1/4
a=0.3
Stretch:
a=60
a=2.5
a=1.1
Step-by-step explanation:
I don't understand what you want solved for here...could you please tell me?
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:
Step-by-step explanation:
By definition, supplementary angles add up to 180 degrees. Therefore, we can set up the follow equation to solve for 
:
Height. Shadow
2.1 m. 5.3 m
X m. 34 m
X=2.1•34/5.3
X=71.4/5.3
X=13.47 m
X~13.5 m
