Answer:
x = 4
Step-by-step explanation:
We can use cross products to solve
84/12 = 28/x
84 * x = 12 *28
Divide each side by 84
84x/84 = 12*28/84
x =4
Answer:
B) 3.2 mi
Step-by-step explanation:
1. The midpoints: midpoint formula = <em>((x₁ + x₂)/2, (y₁+ y₂)/2)</em>
upper = ((6 + -4)/2, (8 + 0)/2) --><em> (1, 4)</em>
lower = ((-8 + 2)/2, (3 + -5)/2) --><em> (-3, -1)</em>
2. Distance: distance formula = <em>√((x₁ - x₂)² + (y₁ - y₂)²)</em>
√((1 + 3)² + (4 + 1)²) = √(16 + 25) = √41 ≈ 6.403... ≈ <em>6.4</em>
3. Scale: 6.4 * 0.5 mi = 3.2 mi
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:
here is a solution :
