There are 2 options to solve that.
1. The first one is by derivatives.
f(x)=x^2+12x+36
f'(x)=2x+12
then you solve that for f'(x)=0
0=2x+12
x=(-6)
you have x so for (-6) solve the first equation, then you find y
y=(-6)^2+12*(-6)+36=(-72)
so the vertex is (-6, -72)
2. The second option is to solve that by equations:
for x we have:
x=(-b)/2a
for that task we have
b=12
a=1
x=(-12)/2=(-6)
you have x so put x into the main equation
y=(-6)^2+12*(-6)+36=(-72)
and we have the same solution: vertex is (-6, -72)
For next task, I will use the second option:
y=x^2-6x
x=(-b)/2a
for that task we have
b=(-6)
a=1
x=(6)/2=3
you have x so put x into the main equation
y=3^2+(-6)*3=(--9)
and we have the same solution: vertex is (3, -9)
Answer:
8
Step-by-step explanation:
The answer is that X=25 because (3x - 15) can be moved in the position where it is a vertical angle because of the corresponding angles postulate
Answer:
Step-by-step explanation:
Answer:
d.
Step-by-step explanation:
if x varies directly as y,
x = ky
8 = 12k
12k = 8
k = 8/ 12
k = 2/3,
x = 2/3 y
now substitute the value of any of the numbers of each pair, and check if it works:
for e.g, find x when y = 15
x = 2/3y
x = 2/3 × 15
x = 10
since you obtain x = 10 when y = 15, the pair is a possible corresponding value of x and y. now do the same for the rest .
ii) x = 2/3y
x = 2/3 × 3, when y = 3
x = 2
hence, this is also a possible pair of x and y values.
iii) x = 2/3y
x = 2/3 × 9 when y = 9
x = 6
again, a possible pair of x and y values.
iv) x = 2/3y
x = 2/3 × 20 when y = 20
x = 40/3
here, when y = 20, x is not 15. it is 40/3.
thus, it can be determined that this is not a possible pair of x and y values.
please try to understand and have a great day!
