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WARRIOR [948]
3 years ago
5

an element with mass 630 grams decays by 18.8% per minute. How much of the element is remaining after 7 minutes, to the nearest

10th of a gram?
Mathematics
2 answers:
Mashutka [201]3 years ago
6 0

Answer:

Step-by-step explanation:

Total Depreciation: 60.24%

Total depreciation: $9,759.50

Car Value after 6 years: $6,440.50

grigory [225]3 years ago
3 0

Answer:

6440.50

Step-by-step explanation:

An element with mass 630 grams decays by 18.8% per minute. How much of the element is remaining after 7 minutes, to the nearest 10th of a gram?

\text{Exponential Functions:}

Exponential Functions:

y=ab^x

y=ab

x

a=\text{starting value = }630

a=starting value = 630

r=\text{rate = }18.8\% = 0.188

r=rate = 18.8%=0.188

\text{Exponential Decay:}

Exponential Decay:

b=1-r=1-0.188=0.812

b=1−r=1−0.188=0.812

\text{Write Exponential Function:}

Write Exponential Function:

y=630(0.812)^x

y=630(0.812)

x

Put it all together

\text{Plug in time for x:}

Plug in time for x:

y=630(0.812)^{7}

y=630(0.812)

7

y= 146.633349

y=146.633349

Evaluate

y\approx 146.6

y≈146.6

round

Your Solution:

6440.50

6440.50

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Answer:

\huge  \: 8 {x}^{3}  \times 3 {x}^{9}  = 24 {x}^{12}

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3 years ago
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ANSWER for the first one:
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8 0
3 years ago
Expand the following using the Binomial Theorem and Pascal’s triangle. (x + 2)6 (x − 4)4 (2x + 3)5 (2x − 3y)4 In the expansion o
ivolga24 [154]
\bf (2x+3)^5\implies 
\begin{array}{llll}
term&coefficient&value\\
-----&-----&-----\\
1&&(2x)^5(+3)^0\\
2&+5&(2x)^4(+3)^1\\
3&+10&(2x)^3(+3)^2\\
4&+10&(2x)^2(+3)^3\\
5&+5&(2x)^1(+3)^4\\
6&+1&(2x)^0(+3)^5
\end{array}

as you can see, the terms exponents, for the first term, starts at highest, 5 in this case, then every element it goes down by 1, till it gets to 0

for the second term, starts at 0, and every element it goes up by 1, till it gets to the highest

now, to get the coefficient, they way I get it, is "the product of the current coefficient and the exponent of the first term, divided by the exponent of the second term plus 1"

notice the first coefficient is always 1

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how did we get 10 for the fourth element?  well, 10*2/4


\bf (2x-3y)^4\implies 
\begin{array}{llll}
term&coefficient&value\\
-----&-----&-----\\
1&&(2x)^4(-3y)^0\\
2&+4&(2x)^3(-3y)^1\\
3&+6&(2x)^2(-3y)^2\\
4&+4&(2x)^1(-3y)^3\\
5&+1&(2x)^0(-3y)^4
\end{array}


\bf (3a+4b)^8\implies 
\begin{array}{llll}
term&coefficient&value\\
-----&-----&-----\\
1&&(3a)^8(+4b)^0\\
2&+8&(3a)^7(+4b)^1\\
3&+28&(3a)^6(+4b)^2\\
4&+56&(3a)^5(+4b)^3\\
5&+70&(3a)^4(+4b)^4\\
6&+56&(3a)^3(+4b)^5\\
7&+28&(3a)^2(+4b)^6\\
8&+8&(3a)^1(+4b)^7\\
9&+1&(3a)^0(+4b)^8
\end{array}

and from there, you can simplify the elements of the expansion by combining the coefficients

like for example, the 7th element of (3a+4b)⁸ will then be 1032192a²b⁶
7 0
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Read 2 more answers
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ANEK [815]

Answer:

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3: 45

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5: 103

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9: 135

10: 45

11: 135

12: 45

13: 77

14: 103

15: 77

16: 103

Step-by-step explanation:

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Ymorist [56]

Answer:

Step-by-step explanation:

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