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Arlecino [84]
3 years ago
9

Reese swam 12 laps in 606 seconds. At that rate, how long would it take Reese to swim 20 laps?

Mathematics
2 answers:
laila [671]3 years ago
5 0

Answer:

1010 seconds to do 20 laps

Step-by-step explanation:

nikklg [1K]3 years ago
3 0

Answer:

1,010 seconds

Step-by-step explanation:

50.5 per lap

50.5 * 20 = 1010

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Could any one help me with this? And with explanation and who ever answers first gets brainliest answer​
ioda

Answer:

gonna solve fore perimeter and area

P = 12 + 3 + 3 + 9 + 9 + 4 + 4 + 4 = 48cm

P = 48 cm

For area or A it is

A = (12*3) + (4*9) = 72 cm^2

Step-by-step explanation:

use deductive reasoning to determine the missing side lengths, and then solve by splitting the figuring into multiple rectangles

5 0
2 years ago
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Evaluate the expression for a = 5, b = 11, and c = 3. 2a + 2c(b − 5) A. 64 B. 46 C. 54 D. 163
Artemon [7]

2a +2c(b-5)

substitute a=5 b=11 c=3

2(5) +2(3) (11-5)

10+6(6)

10+36

46

Choice B


8 0
3 years ago
Solve for the missing angle:
Annette [7]
180 - 47 = 133

The answer is the second option, 133 degrees
8 0
3 years ago
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Find cos (A).<br> Choose 1 answer:<br> A.) 35/12<br> B.) 12/37<br> C.) 35/37<br> D.) 12/35
Mama L [17]

Answer:

c.

Step-by-step explanation:

6 0
3 years ago
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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} &#10;&#10;2 =   e^{ \alpha t} &#10;&#10; \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
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