I did not get any zeros since the graph doesn’t cross the x axis, meaning that there are no rational zeros
However, here is the method u can use to find the zeros lol
You can use the quadratic formula in order to get the zeros
This is the equation therefore use these values
ax^2+bx+c=0
A=1
B= -5
C=12
The quadratic formula is -b±√(b^2-4ac))/2a (I left a picture just in case)
Answer:
Step-by-step explanation:
Given
Required
Solve for s
Add 
to both sides
Take LCM
Answer:
5
Step-by-step explanation:
Answer:
- slant height: 6 units
- lateral area: 108 square units
Step-by-step explanation:
<u>Given</u>
A right regular hexagonal pyramid with ...
- base side length 6 units
- base apothem 3√3 units
- height 3 units
<u>Find</u>
- lateral face slant height
- pyramid lateral surface area
<u>Solution</u>
a) The apothem (a) and height (b) of the pyramid are two legs of the right triangle having the slant height as its hypotenuse (c). The Pythagorean theorem tells us the relationship is ...
c = √(a² +b²) = √((3√3)² +3²) = √(27+9) = √36
c = 6
The slant height of the pyramid is 6 units.
__
b) The lateral surface area of the pyramid is the area of each triangular face, multiplied by the number of faces. The area of one face will be ...
A = (1/2)bh = (1/2)(6 units)(6 units) = 18 units²
Then the lateral surface area is 6 times this value:
SA = 6(18 units²) = 108 units²
The lateral surface area of the pyramid is 108 square units.
