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Anon25 [30]
2 years ago
5

(8.218+9.93)+(17.782+0.07)

Mathematics
1 answer:
Licemer1 [7]2 years ago
5 0
First, using the order of operations(PEMDAS), you would solve what is inside the parenthesis. 
<span>(8.218 + 9.93) + (17.782 + 0.07)
(</span>18.148) + (17.852)
18.148 + 17.852

Now, all you would have to do is add the two sums. 
18.148 + 17.852 = <span>36
</span>
The answer would be 36. 

I hope this helps!

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Keitaro walks at a pace of 3 miles per hour and runs at a pace of 6 miles per hour. Each month, he wants to complete at least 36
Kamila [148]

A. 2x3=6 and 6x12=72 so 78 is greater than 36

6+72=79 which is less than 90

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2 years ago
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A grocery sold 5kg of wheat flour at Rs30 per kg and gained 20%. If he had sold it at Rs27 per kg, what would be his gain or los
maria [59]

Answer:

His gain percent would have been 8%

Step-by-step explanation:

The key to answering this question is to first calculate the price at which the wheat flour was bought.

Mathematically;

% profit = (selling price-cost price)/cost price * 100%

Let the cost price be $x

Thus;

% profit = (30-x)/x * 100

20 = 100(30-x)/x

20x = 3000-100x

100x + 20x = 3000

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x = 3000/120

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So let’s assume he sold at Rs 27

His profit would have been 27-25 = 2

His gain or loss percentage would’ve been;

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3 0
2 years ago
Landon babysits and works part time at tge water park over the summer. onw week, he babysat for 3 hours and worked at the water
Tems11 [23]
B = hourly rate for babysitting and w = hourly rate for working at water park

3b + 10w = 109...multiply by -8
8b + 12w = 177...multiply by 3
----------------------
-24b - 80w = - 872 (result of multiplying by -8)
24b + 36w = 531 (result of multiplying by 3)
---------------------add
- 44w = - 341
w = -341/-44
w = 7.75 <=== hourly rate for working at water park

3b + 10w = 109
3b + 10(7.75) = 109
3b + 77.50 = 109
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8 0
2 years ago
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Neporo4naja [7]

Answer:

0.0198 lbs per gallon

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amount of salt free water = 500 gallons

salt rate in = 1 gal/min

salt rate out = 1 gal/min

amount o salt in brine = 0.25 lb per gallon

Let the amount of salt in the tank be A(t) at any time t.

\frac{dA(t)}{dt} =salt rate in - salt rate out

salt rate in = 0.25 x 1 = 0.25

salt rate out = \frac{A(t)\times 1}{500}

The differential equation is given by

\frac{dA(t)}{dt} =1 - \frac{A(t)\times 1}{500}

where, A(0) = 0

So, the equation becomes

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The solution of the above differential equation is given by

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C(t)=1-e^{\frac{-t}{500}}

Now concentration at t = 10 min

Put, t = 10 min

C(10)=1-e^{\frac{-10}{500}}

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Thus, teh concentration after 10 min is 0.0198 lbs per gallon.

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Answer:

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