vertex = (3,- 5 )
given a quadratic in standard form : y = ax² + bx + c ( a ≠ 0 ), then
the x-coordinate of the vertex is
= - 
y = x² - 6x + 4 is in standard form
with a = 1, b = - 6 and c = 4, hence
= - 
= 3
substitute this value into the equation for y- coordinate
y = 3² - 6(3) + 4 = 9 - 18 + 4 = - 5
vertex = (3, - 5 ) → second table
Hi Jessica,
<span>Remember PEMDAS (Parenthesis, Exponents, Multiplication & Division, Addition & Subtraction).
√3 x 66.15/4.41 {Exponents/Cube & Square Roots First}
1.73 x 66.15 ÷ 4.41 {Multiplication}
114.4395 ÷ 4.41 {Division}
25.95 {Final Answer}
Cheers,
Izzy</span>
So in order to find line AC you must find line AD and DC then plus them together.
to find AD use Pythagoras theorem
a^2 = c^2 - b^2
AD^2 = 7.5^2 - 6.5^2
AD^2 = 56.25 - 42.25
AD^2 = 14
square root both sides to get rid of the ^2
AD ≈ 3.7 or 3.74
Do the same for DC
DC^2 = 10^2 - 6.5^2
DC^2 = 100 - 42.25
DC^2 = 57.75
DC ≈ 7.6
now plus AD and DC which should give u 11.3
Answer:
3/2y -5
Step-by-step explanation:
− 1/2 (−3y+10)
Distribute
-1/2 * -3y + -1/2(10)
3/2y -5
