Yes this is using the transitive property
Point A leads to point B, B goes to C, so overall A goes to C. Think of it connecting a chain.
Nice job on getting the correct answer.
<h3>
Answer: 5</h3>
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Explanation:
Vertex form is
y = a(x-h)^2 + k
We are told the vertex is (3,-2), so we know (h,k) = (3,-2)
y = a(x-h)^2 + k will update to y = a(x-3)^2 - 2
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Then we also know that (x,y) = (4,3) is a point on the parabola. Plug those x and y values into the equation and solve for 'a'
y = a(x-3)^2 - 2
3 = a(4-3)^2 - 2
3 = a(1)^2 - 2
3 = a - 2
3+2 = a
5 = a
a = 5
This is the coefficient of the x^2 term since the standard form is y = ax^2+bx+c.
Let X be the iq score of adults. X is normally distributed with mean of 100 and standard deviation of 20.
If all iq scores are converted to z -scores then we will get z scores. And the z score is standard normal variable with mean=0 and standard deviation =1
Hence the mean and standard deviation of of all z scores will be 0 and 1 respectively.
u have to multiply both of the numbers
Answer:
10.50°C
Step-by-step explanation:
Given x = 2 + t , y = 1 + 1/2t where x and y are measured in centimeters. Also, the temperature function satisfies Tx(2, 2) = 9 and Ty(2, 2) = 3
The rate of change in temperature of the bug path can be expressed using the composite formula:
dT/dt = Tx(dx/dt) + Ty(dy/dt)
If x = 2+t; dx/dt = 1
If y = 1+12t; dy/dt = 1/2
Substituting the parameters gotten into dT/dt we will have;
dT/dt = 9(1)+3(1/2)
dT/dt = 9+1.5
dT/dt = 10.50°C/s
Hence the rate at which the temperature is rising along the bug's path is 10.50°C/s
