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Pie
3 years ago
11

Sara had some candy in her pocket. She first kept 2 pieces herself and then gave her 5 children 3 pieces each. If she didn't hav

e any candy left over, how many pieces of candy did she start with
Mathematics
1 answer:
Juli2301 [7.4K]3 years ago
5 0

17 pieces. She had two and she had to have 15 to give to the children. If each child got 3 and there is 5. This turns into 3*5=15. Then you would add 15+2 to get 17

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The radius of a circle is 2.6 in. Find the circumference <br> to the nearest tenth.
Georgia [21]

Answer:

16.

Step-by-step explanation:

since given the radius and the formula of the circumference of a circle is 2pie*r

5 0
3 years ago
1200$ invested at a rate of 3.5% compounded quarterly; 4 years
Tju [1.3M]

Step-by-step explanation:

<em>1st year</em>:

(1200 x 3.5 x 1) ÷ 100 = $42

<em>2nd year:</em>

(1242 x 3.5 x 1) ÷ 100= $43.47

<em>3rd year:</em>

(1285.47 x 3.5 x 1) ÷ 100= $44.99≈ $45

<em>4th year:</em>

(1330.47 x 3.5 x 1) ÷ 100= $46.56

Compound interest:

$(42 + 43.47 + 45 + 46.56)

=<u>$ 177.03</u>

4 0
3 years ago
Pls pls help me I will mark you brainest
Musya8 [376]

Answer:

It would be 125m^2

Step-by-step explanation:

17*10=170

170/2=85

8*10=80

80/2=40

85+40=125

7 0
2 years ago
Read 2 more answers
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




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3 years ago
11/12 as a percentage
evablogger [386]

Answer:

91.67%

Step-by-step explanation:

8 0
1 year ago
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