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Vesna [10]
2 years ago
13

Mario is baking pls help

Mathematics
2 answers:
sattari [20]2 years ago
7 0
3/4 is the same as 6/8. So 6/8 plus 5/8 is 11/8. But you can’t have a numerator greater than the denominator so that is equal to 1 3/8. Then you add that to the 15 and you get 16 3/8. Hope this helps!!!
agasfer [191]2 years ago
3 0

Answer:

b

Step-by-step explanation:

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The value of a motorbike depreciates by 32% per year. Work out the current value of a motorbike bought 5 years ago for £3600
DaniilM [7]
\bf \qquad \textit{Amount for Exponential Decay}
\\\\
A=P(1 - r)^t\qquad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{initial amount}\to &3600\\
r=rate\to 32\%\to \frac{32}{100}\to &0.32\\
t=\textit{elapsed time}\to &5\\
\end{cases}
\\\\\\
A=3600(1-0.32)^5\implies A=3600(0.68)^5\implies A\approx 523.41601
3 0
3 years ago
Please help! It will be appreciated!
AysviL [449]
It’s d, looks like you got it already! With ratios, just pay attention to not only the numbers but order of the words as well, so you can be sure the ratio follows the problem.
8 0
3 years ago
Read 2 more answers
PLEASE HELP ME !!!!!!!
Kruka [31]

Answer:

63*42=2646

22*22=484

Area of Shaded Gray portion=2646-484=2162

7 0
2 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} 

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
$3.98 divided by 100x4
Oliga [24]
$0.00995 is the correct answer.

$3.98/400=0.00995
4 0
3 years ago
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