Answer:
If Anne’s age is represented by the variable a then express Jenny’s age in terms of a if Jenny is three years less than half of Anne’s age
Because both of the equations = s, we can set them equal to each other.
c + 7 = 2c - 13. Now we can solve for c. Once you have found c, you can substitute it back into either of the two original equations and solve for s.
Answer: Choice A
f(x) = (x + 2)^2 - 11
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Work Shown:
y = x^2 + 4x - 7
y+7 = x^2 + 4x
y+7+4 = x^2 + 4x + 4
y+11 = x^2 + 4x + 4
y+11 = (x + 2)^2
y = (x + 2)^2 - 11
f(x) = (x + 2)^2 - 11
In the third step, I added 4 to both sides to complete the square for the x^2+4x portion. Notice that (x+2)^2 = x^2+4x+4. So I added 4 to fill in the missing piece needed to complete the square.
Put another way the '4' added to both sides is because we first divided the x coefficient 4 in half to get 4/2 = 2. Then you square it to get 2^2 = 4.
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Alternative Method (optional)
y = x^2 + 4x - 7 is in the form of y = ax^2+bx+c
where: a = 1, b = 4, c = -7
Plug those a,b values into the formula below
h = -b/(2a)
h = -4/(2(1))
h = -2
This is the x coordinate of the vertex.
Use it to find the y coordinate of the vertex.
y = x^2 + 4x - 7
y = (-2)^2 + 4(-2) - 7
y = -11
The vertex is located at (h,k) = (-2,-11)
We have the template y = a(x-h)^2 + k update to y = (x + 2)^2 - 11 after plugging in a = 1, h = -2, and k = -11.
Answer:
Step-by-step explanation:
Let X denote the random variable that obeys the normal distribution.
Given that:
<u>For morning flights</u>
Mean = 15
standard deviation = 5
Sample size = 10
The Z - score is calculated as:
Z = 3
<u>For evening flights</u>
Mean = 20
standard deviation = 3
Sample size = 10
Z = 3.33
Hence, from the above z-scores, we will realize that the evening flight is more late than usual.