Answer: No mistakes was made
Step-by-step explanation:
Because It all equals the same as the answer
Let that be
Two vertical asymptotes at -1 and 0
If we simply
- Denominator has degree 2
- Numerator should have degree as 2 and coefficient as 3 inorder to get horizontal asymptote y=3 means the quadratic equation should contain 3x²
- But there should be a x intercept at -3 so one zeros should be -3
Find a equation
Find zeros
Horizontal asymptote
So our equation is
Graph attached
Answer:
intersection point (-2,5)
Step-by-step explanation:
the line y = - 5 / 2 x you do the reciprocal which gets you x = -2 y = 5
Answer:
126
Step-by-step explanation:
Let x be the missing length
The triangles are similar:
● UE/140 = 45/x
From the graph we deduce that:
● UE = 140 - 90 = 50
Replace UE by its value
● 50/ 140 = 45/x
Switch x and 50
● x / 140 = 45/50
45/50 is 9/10 wich is 0.9
● x/140 = 0.9
Multiply 0.9 by 140
● x = 140 × 0.9
● x = 126
6.
4x^2 + 4 = 0
Divide both sides by 4
x^2 + 1 = 0
Use the quadratic formula since this cannot be factored.
x = (-b +- sqrt(b^2 - 4ac))/(2a)
x = +- sqrt(-4(1)(1))/2
x = +- sqrt(-4)/2
x = +- 2i/2
x = +- i
x = i or x = -i
Quicker solution:
If you have x^2 = number, then
x = +- sqrt(number)
Once you get to
x^2 + 1 = 0
Subtract 1 from both sides
x^2 = -1
Apply the quick method
x = +- sqrt(-1)
x = +- i
8.
2x^2 + 50 = 0
Divide both sides by 2
x^2 + 25 = 0
Subtract 25 from both sides
x^2 = -25
Apply quick method
x = +- sqrt(25)
x = +- 5i
x = 5i or x = -5i
