Hello!
You can make two expressions based on what you know
let x equal the larger number
let y equal the smaller number
x + y = 95
x + 2y = 120
We can use elimination to get rid of x
You can subtract the two equation to get rid of x
-y = -25
Since y is negative we can multiply both sides by -1 to make y positive
y = 25
Now we can put this into one of the other equations to find x
x + 25 = 95
subtract 95 from both sides
x = 70
The answers are 70 and 25
Hope this helps!
Multiply:
6•12=72
Then, you subtract 39 from 72:
72-39=33
33 containers would still need to be set up.
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:
A. drag it to the right
b. drag it to the left
Step-by-step explanation:
