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scoundrel [369]
3 years ago
6

An element with mass 310 grams decays by 8. 9% per minute. How much of the element is remaining after 19 minutes, to the nearest

10th of a gram?.
Mathematics
1 answer:
Cerrena [4.2K]3 years ago
3 0

\qquad \textit{Amount for Exponential Decay} \\\\ A=P(1 - r)^t\qquad \begin{cases} A=\textit{current amount}\\ P=\textit{initial amount}\dotfill &310\\ r=rate\to 8.9\%\to \frac{8.9}{100}\dotfill &0.089\\ t=\textit{elapsed time}\dotfill &19\\ \end{cases} \\\\\\ A=310(1-0.089)^{19}\implies A=310(0.911)^{19}\implies A\approx 52.8

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I need help i will give the brainleness thing to whoever can answer
dimaraw [331]

Answer:

A=30cm, B=39cm

Step-by-step explanation:

They are an enlargement. Therefore, the sides are larger by the same proportion, otherwise, it wouldn't be an enlargement. Therefore, the bottom side of the big one is 3 times that of the small one. Therefore, the other sides are also 3 times, making A = 30cm and B = 39cm.

8 0
3 years ago
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Please help quick I will give brainliest :(
Flura [38]

Answer:

r=4 d=8 a = 16π c = 8π

d=6 r=3 a=9π c=6π

a=36π r=6 d=12 c=12π

c=18π r=9 d=18 a=81π

Step-by-step explanation:

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6 0
3 years ago
Here this is the new one.
Helga [31]

Formula #1: 0.15x + 22

Formula #2: 0.19x

0.15x + 22 = 0.19x

15x+2200=19x\\15x=19x-2200\\-4x=-2200\\\frac{-4x}{-4}=\frac{-2200}{-4}\\X = 550

So x = 550.

I hope this helps! :)

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X+2/x–2 –x–2/x+2 = 5/6<br>​
Anna71 [15]
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3 years ago
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Grade 5 farm has 85 rows of cabbages.If there are 58 cabbages in each row ,how many cabbages are there in the farm to the neares
ElenaW [278]

The number of cabbages on the farm to the nearest thousand is 5000 based on the given parameters

<h3>How to determine the number of cabbages on the farm to the nearest thousand</h3>

The number of rows is given as:

Number of rows = 85

The number of cabbage on each row is

Number of Cabbage on each row = 58

The number of cabbages on the farm is calculated as:

Farm = Number of rows * Number of Cabbage on each row

Substitute the known values (i.e. Number of rows = 85 and Number of Cabbage on each row = 58) in the above equation

Farm = 85 * 58

Evaluate the product of 85 and 58

Farm = 4930

Approximate the above result (i.e. 4930) to the nearest thousand

Farm = 5000

Based on the given parameters, the number of cabbages on the farm to the nearest thousand is 5000

So, the complete parameters are:

Number of rows = 85

Number of Cabbage on each row = 58

Number of Cabbage in the farm = 5000

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