Answer:
8.6%
Step-by-step explanation:
To find the percent change, you will need to compute the positive difference and then divide the difference by the original (the older amount).
So the positive difference will be obtain by doing larger minus smaller:
6300
- 5800
-----------
500
The older amount was 5800.
So 500/5800 is the answer as a un-reduced fraction.
I'm going to reduce it by dividing top and bottom by 100:
500/5800 = 5/58
5/58 is the answer as a reduced fraction.
5 divided by 58 gives=0.086206897 in the calculator .
Approximately 0.0862 is the answer as a decimal.
To convert this to a percentage, multiply it by a 100:
8.62%
Rounded to the nearest tenths is 8.6%
-------------
So 5800+5800(.0862) should be pretty close to 6300 (not exactly though since we rounded).
5800+5800(.0862)=6299.96 using the calculator.
9514 1404 393
Answer:
axis of symmetry: x = 2
see below for a graph
Step-by-step explanation:
You can complete the square to put the equation in vertex form.
y = (x^2 -4x +4) -1 -4
y = (x -2)^2 -5 . . . . . . vertex is (2, -5)
The equation of the axis of symmetry is the equation of the vertical line through the vertex:
x = 2 . . . . . axis of symmetry
__
Points on the graph are easily found by using integer values on either side of x=2. Points on one side of the axis of symmetry are mirrored on the other side of it.
A picture types of the models would be helpful.. but they could be found in a rectangular shape or square such as width=2 and length =12 vise versa or w=4 and length = 6 , etc etc
Answer: t-half = ln(2) / λ ≈ 0.693 / λExplanation:The question is incomplete, so I did some research and found the complete question in internet.
The complete question is:
Suppose a radioactive sample initially contains
N0unstable nuclei. These nuclei will decay into stable
nuclei, and as they do, the number of unstable nuclei that remain,
N(t), will decrease with time. Although there is
no way for us to predict exactly when any one nucleus will decay,
we can write down an expression for the total number of unstable
nuclei that remain after a time t:
N(t)=No e−λt,
where λ is known as the decay constant. Note
that at t=0, N(t)=No, the
original number of unstable nuclei. N(t)
decreases exponentially with time, and as t approaches
infinity, the number of unstable nuclei that remain approaches
zero.
Part (A) Since at t=0,
N(t)=No, and at t=∞,
N(t)=0, there must be some time between zero and
infinity at which exactly half of the original number of nuclei
remain. Find an expression for this time, t half.
Express your answer in terms of N0 and/or
λ.
Answer:
1) Equation given:
← I used α instead of λ just for editing facility..
Where No is the initial number of nuclei.
2) Half of the initial number of nuclei:
N (t-half) = No / 2So, replace in the given equation:
3) Solving for α (remember α is λ)
αt ≈ 0.693
⇒ t = ln (2) / α ≈ 0.693 / α ← final answer when you change α for λ