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ale4655 [162]
3 years ago
14

Can you please help me with this? tytytyty

Mathematics
1 answer:
Ray Of Light [21]3 years ago
3 0
Deletion: CATTCACG

insertion: CATTTCACACG

inversion: CATTGCACAC 

duplication: CATTCACACCACG

substitution: CATTCACACA
You might be interested in
What are the measures of angles M and N?
romanna [79]

Answer:

mM = 113 and mN = 61

Step-by-step explanation:

In a Cyclic quadrilateral, the rule states that:

The sum opposite interior angles is equal to 180°

In the above diagram, we have cyclic quadrilateral KLMN

According to the rule stated above:

Angle K is Opposite to Angle M

So, Angle K + Angle M = 180°

Angle K = 67°

67° + Angle M = 180°

Angle M = 180° - 67°

Angle M = 113°

Angle L is Opposite to Angle N

so Angle L + Angle N = 180°

Angle L is given as = 119°

119° + Angle N = 180°

Angle N = 180° - 119°

Angle N = 61°

Therefore, Angle M = 113° and Angle N = 61°

7 0
3 years ago
Please help it’s lines
JulijaS [17]
I think the answer is c, 88
Hope this helps :)
6 0
3 years ago
Since at t=0, n(t)=n0, and at t=∞, n(t)=0, there must be some time between zero and infinity at which exactly half of the origin
Airida [17]
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:

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:

N(t)=N _{0} e^{-  \alpha  t} ← 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:

N_{t-half} =  N_{0} /2 =  N_{0}  e^{- \alpha t}

3) Solving for α (remember α is λ)

\frac{1}{2} =  e^{- \alpha t} 

2 =   e^{ \alpha t} 

 \alpha t = ln(2)

αt ≈ 0.693

⇒ t = ln (2) / α ≈ 0.693 / α ← final answer when you change α for λ




4 0
3 years ago
Pls help me find the answer
gregori [183]
The surface area is 1536
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4 years ago
Paul is planning to extend the bakery and is looking at the plans. The length of the shortest wall is 17 meters
Luda [366]
About 2.9411 I believe
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