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Snezhnost [94]
3 years ago
15

I don't get this help I'm in 5th :(

Mathematics
2 answers:
monitta3 years ago
8 0

Answer:

46m x 2.5m= 115m^{2}  for the shaded region

Step-by-step explanation:

46 x 5= 230m^{2}

230m^{2}÷ 2 = 115m^{2} for the shaded region

        or

46m × 2.5m = 115m^{2}  for the shaded region

Degger [83]3 years ago
5 0
I don't think Hailey is correct because 5 is half and you need to multiply the half.
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At the school the ratio of boys to girls is 5:3 if there are 385 boys in the school, how many students are there altogether?
ser-zykov [4K]

Answer:

There are 616 total students

Step-by-step explanation:

boys : girls: total

5           3      5+3 = 8

There are 385 boys

385/5 = 77

Multiply each term by 77

boys : girls: total

5*77  3*77   8*77

385    539    616

There are 616 total students

6 0
2 years ago
Read 2 more answers
Find rate of change help show work.
sukhopar [10]

Answer:

rise/run = 2/-4 or it is rise/run = -2/4

3 0
2 years ago
In circle c with diameter SP, what is the measure of PCQ?
Eva8 [605]

Answer:

i dont know

Step-by-step explanation:

3 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
2 years ago
8th grade math help please zoom in on the question for a better look (will give brainliest and thanks if it's right)
eduard
The two pairs are Pythagorean triples because if you plug the two legs of a right triangle into the Pythagorean theorem(A^2+B^2=C^2), then you will find the measurement for the third side(hypotenuse). i.e. 15^2+12^2=9^2(it's the Pythagorean triple 3,4,5 multiplied by 3). This works with any triple, as long as your using the legs of the triangle and as long as the triangle is a right triangle.
5 0
2 years ago
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