Answer:
see below
Step-by-step explanation: 7 1 16 33
y = x² translated t the point (3, 2) y = 0 when x = 0
y = (x - 3)² moves the function three units to the right y = 0 when x = -3
y = (x-3)² + 2 moves the function up 2 units y = 2 when x = -3
y = (x-3)² + 2
y = x² - 6x + 9 + 2
y = x² - 6x +11
graph the equations x², (x-3)² + 2, and x² - 6x + 11 to verify (I did)
Answer:
4.837
Step-by-step explanation:
-2 + 3b = 5b - 2(2b + 3) Distribute the -2 through the parentheses.
-2 + 3b = 5b - 4b - 6 Combine like terms
-2 + 3b = b - 6 Bring all "b's" to one side of the equation
-2 + 3b - b = b - b - 6 Combine like terms
-2 + 2b = -6 Bring the 2 to the right side of the equation
-2 + 2 + 2b = -6 + 2 Combine like terms
2b = -4 Isolate b by dividing both sides by 2
b = -4/2 = -2
Answer b = -2
Answer:
464 grams.
Step-by-step explanation:
Amount of substance:
The amount of substance after t years is given by an equation in the following format:

In which A(0) is the initial amount and r is the decay rate, as a decimal.
Co-60 has a half life of 5.3 years.
This means that:

We use this to find r. So



![\sqrt[5.3]{(1-r)^{5.3}} = \sqrt[5.3]{0.5}](https://tex.z-dn.net/?f=%5Csqrt%5B5.3%5D%7B%281-r%29%5E%7B5.3%7D%7D%20%3D%20%5Csqrt%5B5.3%5D%7B0.5%7D)


So

If a pellet that has been in storage for 26.5 years contains 14.5g of Co-60, how much of this radioisotope was present when the pellet was put in storage?
We have that
, and use this to find A(0). So



[tex]A(0) = 464[tex]
464 grams.