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jasenka [17]
3 years ago
8

3/8 + 1/8 - 2/7 + 1/4

Mathematics
2 answers:
antiseptic1488 [7]3 years ago
8 0

Answer:

0.46428571428

Step-by-step explanation:

Nimfa-mama [501]3 years ago
7 0

Answer:

<h2>13/28</h2>

Step-by-step explanation:

3/8 + 1/8 = <em>4/8</em> = <em>1/2 </em>= <em>0.5</em>

1/2 - 2/7 = <em>3/14</em> = <em>0.214285714286</em>

3/14 + 1/4 = <em>13/28 </em>= <em>0.464285714286</em>

<h2><em>13/28 or 0.464285714286</em></h2><h2 />

<em>Hope this helps! <3</em>

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I need to know if it’s A B C OR D
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SOLUTION

Consider the property, opposite angle of a cyclic quadrilateral are supplementary

hence

\angle NMP+\angle NOP=180^0

Recall that

\begin{gathered} \angle NMP=\angle M=(8x-24)^0 \\ \text{and } \\ \angle NOP=\angle O=(4x)^0 \end{gathered}

Hnece, we have that

\begin{gathered} \angle M+\angle O=180^0(opposite\text{ angles of a cyclic quadrilateral)} \\ (8x-24)^0+4x^0=180^0 \end{gathered}

Then

\begin{gathered} 8x-24+4x=180 \\ 8x+4x-24=180 \\ 12x-24=180 \\ \text{Add 24 t o both sides } \\ 12x-24+24=180+24 \\ 12x=204 \\ \text{divide both sides by 12} \\ \frac{12x}{12}=\frac{204}{12} \\ \text{Then} \\ x=17 \end{gathered}

Hence

x=17

Since

\begin{gathered} \angle NOP=(4x)^0 \\ \text{substitute the value of x, we have } \\ \angle NOP=4\times17=68^0 \\ \text{Hence } \\ \angle NOP=68^0 \end{gathered}

Therefore

The measure of angle NOP is 68⁰

Answer; 68⁰ (The fourth Option )

5 0
1 year ago
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Answer:

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Step-by-step explanation:

the result becomes negative because 7.273 is less than 12.37

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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
3/30 = n/50<br><br> n = (Blank)<br><br> Find n.
UNO [17]
<span>3/30 = n/50
</span><span>1/10 = n/50
10n = 50
n = 50/10
n=5</span>
3 0
3 years ago
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