A number that is a multiple of both 2 and 5 is 10
Answer:
False
False
True
False
True
Step-by-step explanation:
I hope this helps!
Answer:

Step-by-step explanation:
Given
Shape: Opening of a circular vase


Required
Determine the area
This opening implies a circle; hence the area will be calculated using area of a circle.

Solve for radius (r)



Substitute 6.75 for r in 


--- Approximated
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