Answer:
x=12
Step-by-step explanation:
That's the right answer
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:
Into 1/3 pieces
Step-by-step explanation:
1/3 + 1/3 + 1/3 = 1 the whole pizza pie, yum
If c = 8 and d = -5:
a) c - 3 = 8 - 3
= 5
b) 15 - c = 15 - 8
= 7
c) 3(c + d) = 3(8 + (-5))
= 3*3
= 9
d) 2c - 4d = 2(8) - 4(-5)
= 16 + 20
= 36
e) d - c^2 = -5 - (8)^2
= -5 - 64
= -69
f) 2d^2 + 5d = 2(-5)^2 + 5(-5)
= 50 - 25
= 25