Answer:
Step-by-step explanation:
If I remember expanded form is putting all number in terms of being times 10 to a power. starting with the ones place it's x10^0 which of course is just 1, so you don't need to include the x10 at all.
now, moving one either way that 0 either get 1 added to it or subtracted. So in the 10s place it becomes x10^1 and the tenths place becomes x10^-1. Similarly the hundreds place and hundreths place will have x10^2 and x10^-2 respectively. So keep this pattern going to find each place of the number. Can you list out what places are displayed in 5.625?
I will get you started, but let me know if you don't quite get it.
5 + 6*10^-1 + ...
Can you figure out the rest?
Answer:
I KNOWW THE ANSWER!!
Step-by-step explanation:
Answer:
11.6 miles
Step-by-step explanation:
Given that:
Distance from J2 to J3 = 8.7 miles
Distance between J1 and J2 = 1/3 of J2 toJ3
Hence, distance between J1 and J2 becomes :
1/3 * 8.7 miles = 2.9 miles
Distance from J1 to J3 equals :
(J1 to J2) + (J2 to J3)
2.9 miles + 8.7 miles
= 11.6 miles
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.
Unfortunately, there is no way I can find the value of X and Y because there is no question for it.
You can either find the value of X and Y in the problem or make X and Y your own value if you need to find something that is equivalent.
I hope this helped.
