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mina [271]
2 years ago
13

Find the perimeter of the rectangle whose vertices are the points with coordinates (0,0), (2,0), (0,5), and (2,5).

Mathematics
1 answer:
zheka24 [161]2 years ago
3 0

Answer: The perimeter is 14 units

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Angle 0 lies in the second quadrant and sin 0 = 3/5 cos= coy =
Elza [17]
Consider the right triangle ABC with legs AB=4, AC=3 and hypotenuse BC=5. Angle B has
 \sin B=\frac{opposite}{hypotenuse} = \frac{3}{5} and
 \cos B = \frac{adjacent}{hypotenuse} = \frac{4}{5}.

Since O lies in second quadrant \cos O\ \textless \ 0 and 
\cos O= -\cos B =- \frac{4}{5}.
Answer: \cos O=- \frac{4}{5}. 


3 0
3 years ago
Please do this correct if inappropriate then report and no LINKS and also will do brainly
Nuetrik [128]
Mean=87.5
Median=88
Mode=79,85,86,90,92,93 so technically all the numbers show are the mode cause they all repeat 1 time but if there were two of the same number that would be the mode cause there’s two and not one so you always go by which one repeats more!
4 0
2 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
A bakery sells pastries in boxes of 7 each. If the bakery has 27 pastries, how many boxes can be filled?​
skad [1K]

Answer:

3 with a few left over so 4

Step-by-step explanation:

4 0
3 years ago
The table contains data on the number of people visiting a historical landmark over a period of one week.
Natasha2012 [34]

Answer:

A .........

I think its the right answer

haha

3 0
3 years ago
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