A balanced coin will have 50:50 chance of showing either head or tail. In this question, you are asked the probability to find the first head after two tosses. That means the toss result should be:
1. tail = 1/2 probability
2. head = 1/2 probability
Previously you are tossing the coin 10 times and obtain zero head. But this previous 10 tosses result should not influence your probability in the 11th and 12th toss. The chance will always be 50:50 no matter how much the tosses count. So the probability should be: 1/2 * 1/2= 1/4
1 out of 6 so 1/6. there are 6 numbers and you only have a one out of 6 chance to roll that specific number (7).
Answer C (-8,0)
because the line crosses the x-intercept (the horizontal line)
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)
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