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Pie
2 years ago
10

An element with mass 730 grams decays by 28.8% per minute. How much of the element is remaining after 10 minutes, to the nearest

10th of a gram?
Mathematics
2 answers:
Lana71 [14]2 years ago
8 0

Answer:

A= 24.4 grams

Step-by-step explanation:

An element with mass 730 grams decays by 28.8% per minute

For exponential decay we use formula

A= P(1-r)^t

Where P is the initial amount of element present

A is the amount remaining

r is the decay rate

and t is the time period in minutes

P= 730 grams, r= 28.8%= 28.8/100= 0.288, t= 10 minutes

A= 730(1-0.288)^10

A=24.44122

Round to nearest tenth

So A= 24.4 grams

trasher [3.6K]2 years ago
6 0
\bf \qquad \textit{Amount for Exponential Decay}\\\\
A=P(1 - r)^t\qquad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{initial amount}\to &730\\
r=rate\to 28.8\%\to \frac{28.8}{100}\to &0.288\\
t=\textit{elapsed time}\to &10\\
\end{cases}
\\\\\\
A=730(1-0.288)^{10}\implies A=730(0.712)^{10}
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Answer:

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(b)Therefore, the length of the fence is  97\frac{1}{2}$ feet

Step-by-step explanation:

The diagram of the yard is attached below.

(a)I have labeled the missing dimensions of the yard as x and y.

Therefore:

y+8\frac{1}{4}=17 \frac{1}{4}\\y=17 \frac{1}{4}-8\frac{1}{4}\\y=9$ feet

Similarly:

x+15\frac{1}{4}=31\frac{1}{2}\\x=31\frac{1}{2}-15\frac{1}{4}\\x=31-15+\frac{1}{2}-\frac{1}{4}\\x=16+\frac{1}{4}\\x=16\frac{1}{4}$ feet

The missing measurements are 9 feet and 16\frac{1}{4}$ feet.

(b)Length of the Fence

The fence is rectangular shaped with:

Length = 31\frac{1}{2}$ feet

Width = 17 \frac{1}{4}$ feet

Perimeter of a Rectangle = 2(L+W)

Therefore, the length of the fence

=2(31\frac{1}{2}+17 \frac{1}{4})\\=2(31+17+\frac{1}{2}+ \frac{1}{4})\\=2(48+ \frac{3}{4})\\=96+\frac{3}{2}\\=97\frac{1}{2}$ feet

8 0
2 years ago
The probability of drawing two aces from a standard deck is 0.0059. We know this probability, but we don't know if the first car
ivanzaharov [21]

Answer:

Option C is right

C. They are independent because, based on the probability, the first ace was replaced before drawing the second ace.

Step-by-step explanation:

Given that the  probability of drawing two aces from a standard deck is 0.0059

If first card is drawn and replaced then this probability would change.  By making draws with replacement we make each event independent of the other

Drawing ace in I draw has probability equal to 4/52, when we replace the I card again drawing age has probability equal to same 4/52

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4 0
1 year ago
Oleg, Sasha, and Dima share 600 toys. Sasha has twice as many toys as Oleg. Dima has 40 more toys than Oleg. How many toys does
Anika [276]

Answer: 140

Oleg, Sasha, and Dima shared 600 toys. Sasha had twice as many toys as Oleg. Dima had 40 more toys than Oleg.

Let the number of toys Oleg has = x

Let the number of toys Sasha has = y

Let the number of toys Dima has = z

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So y = 2*x= 2x

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z = x + 40

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Step-by-step explanation:

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2 years ago
the table shows the number of minutes Tim has for lunch and study hall. he calculates that these two periods account for 18% of
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Tim spends 500 minutes at the school.

Step-by-step explanation:

Given,

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Let,

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Tim spends 500 minutes at the school.

Keywords: percentage, division

Learn more about division at:

  • brainly.com/question/101683
  • brainly.com/question/103144

#LearnwithBrainly

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Solve for x in the equation 2x^2+3x-7=x^2+5x+39
Shalnov [3]
Hey there, hope I can help!

\mathrm{Subtract\:}x^2+5x+39\mathrm{\:from\:both\:sides}
2x^2+3x-7-\left(x^2+5x+39\right)=x^2+5x+39-\left(x^2+5x+39\right)

Assuming you know how to simplify this, I will not show the steps but can add them later on upon request
x^2-2x-46=0

Lets use the quadratic formula now
\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}
x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

\mathrm{For\:} a=1,\:b=-2,\:c=-46: x_{1,\:2}=\frac{-\left(-2\right)\pm \sqrt{\left(-2\right)^2-4\cdot \:1\left(-46\right)}}{2\cdot \:1}

\frac{-\left(-2\right)+\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)}}{2\cdot \:1} \ \textgreater \  \mathrm{Apply\:rule}\:-\left(-a\right)=a \ \textgreater \  \frac{2+\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)}}{2\cdot \:1}

Multiply the numbers 2 * 1 = 2
\frac{2+\sqrt{\left(-2\right)^2-\left(-46\right)\cdot \:1\cdot \:4}}{2}

2+\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)} \ \textgreater \  \sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)}

\mathrm{Apply\:rule}\:-\left(-a\right)=a \ \textgreater \  \sqrt{\left(-2\right)^2+1\cdot \:4\cdot \:46} \ \textgreater \  \left(-2\right)^2=2^2, 2^2 = 4

\mathrm{Multiply\:the\:numbers:}\:4\cdot \:1\cdot \:46=184 \ \textgreater \  \sqrt{4+184} \ \textgreater \  \sqrt{188} \ \textgreater \  2 + \sqrt{188}
\frac{2+\sqrt{188}}{2} \ \textgreater \  Prime\;factorize\;188 \ \textgreater \  2^2\cdot \:47 \ \textgreater \  \sqrt{2^2\cdot \:47}

\mathrm{Apply\:radical\:rule}: \sqrt[n]{ab}=\sqrt[n]{a}\sqrt[n]{b} \ \textgreater \  \sqrt{47}\sqrt{2^2}

\mathrm{Apply\:radical\:rule}: \sqrt[n]{a^n}=a \ \textgreater \  \sqrt{2^2}=2 \ \textgreater \  2\sqrt{47} \ \textgreater \  \frac{2+2\sqrt{47}}{2}

Factor\;2+2\sqrt{47} \ \textgreater \  Rewrite\;as\;1\cdot \:2+2\sqrt{47}
\mathrm{Factor\:out\:common\:term\:}2 \ \textgreater \  2\left(1+\sqrt{47}\right) \ \textgreater \  \frac{2\left(1+\sqrt{47}\right)}{2}

\mathrm{Divide\:the\:numbers:}\:\frac{2}{2}=1 \ \textgreater \  1+\sqrt{47}

Moving on, I will do the second part excluding the extra details that I had shown previously as from the first portion of the quadratic you can easily see what to do for the second part.

\frac{-\left(-2\right)-\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)}}{2\cdot \:1} \ \textgreater \  \mathrm{Apply\:rule}\:-\left(-a\right)=a \ \textgreater \  \frac{2-\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)}}{2\cdot \:1}

\frac{2-\sqrt{\left(-2\right)^2-\left(-46\right)\cdot \:1\cdot \:4}}{2}

2-\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)} \ \textgreater \  2-\sqrt{188} \ \textgreater \  \frac{2-\sqrt{188}}{2}

\sqrt{188} = 2\sqrt{47} \ \textgreater \  \frac{2-2\sqrt{47}}{2}

2-2\sqrt{47} \ \textgreater \  2\left(1-\sqrt{47}\right) \ \textgreater \  \frac{2\left(1-\sqrt{47}\right)}{2} \ \textgreater \  1-\sqrt{47}

Therefore our final solutions are
x=1+\sqrt{47},\:x=1-\sqrt{47}

Hope this helps!
8 0
1 year ago
Read 2 more answers
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