Answer:
Given that,
The number of grams A of a certain radioactive substance present at time, in years
from the present, t is given by the formula
a) To find the initial amount of this substance
At t=0, we get
We know that e^0=1 ( anything to the power zero is 1)
we get,
The initial amount of the substance is 45 grams
b)To find thehalf-life of this substance
To find t when the substance becames half the amount.
A=45/2
Substitute this we get,
Taking natural logarithm on both sides we get,
Half-life of this substance is 154.02
c) To find the amount of substance will be present around in 2500 years
Put t=2500
we get,
The amount of substance will be present around in 2500 years is 0.000585 grams
Answer: Yes, y does vary directly with x.
Constant of variation = 1/4
The function rule is y = (1/4)x
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Explanation:
Let's assume that y does vary directly with x.
If that's the case, then we have an equation in the form y = kx, where k is the constant of variation.
Solving for k gets us k = y/x
For each row, divide the y value over the x value
- row one: k = y/x = 14/56 = 0.25
- row two: k = y/x = 20/80 = 0.25
- row three: k = y/x = 22/88 = 0.25
Each row yields the value k = 0.25 and it fully confirms y does vary directly with x.
So y = kx becomes y = 0.25x as the function rule, which is equivalent to y = (1/4)x
Answer:
A(3) = -6 + (3 - 1) (5)
-6 + (2)(5)
-6 + (10)
4
A(4) = -6 + (4 - 1) (5)
-6 + (3)(5)
-6 + 15
9
A(10) = -6 + (10 - 1) (5)
-6 + (9)(5)
-6 + 45
39
Reorder both sides
3y-6x=9
7-4y-3x=9
Multiply both sides
-6x-3y=9
6x+8y=-4
Dived both sides
11y= 5
Y=5/11
Substitute the value of Y
-3x-4(5/11)
Equals —> -14/11
Answer : (5/11 , -14/11)
