Circumfrence: pi times diameter
7.9*
circumfrence is approximatley 24.806
area:
a=*(1/2(d))^2
a= 48.99
7.9*
circumfrence is approximatley 24.806
area:
a=
a= 48.99
2+2 Please help I’ve been try this all day long
2 answers:
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volume of the box is 675 cubic inches
A machine produces open boxes using square sheets of plastic.
It is a square sheet so length and width are same
Lets assume length as x so width is also x
The machine cuts equal-sized squares measuring 3 inches on a side from each corner of the sheet.
After turning up the sides the height of the box becomes 3 inches
We know the volume of a box formula
Volume = Length * width * height
We know length is x , width is x and height = 3
So V = x * x * 3
Given volume = 675 cubic inches
Divide by 3 on both sides
Now we take square root on both sides
x = 15
the length of one side of the open box is 15 inches.
Answer: t-half = ln(2) / λ ≈ 0.693 / λ
Explanation:
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:
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 / 2
So, 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
Explanation:
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:
Where No is the initial number of nuclei.
2) Half of the initial number of nuclei: N (t-half) = No / 2
So, 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