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mafiozo [28]
4 years ago
5

.37 7 repeating as a fraction

Mathematics
1 answer:
Wittaler [7]4 years ago
8 0
0.37777... = 3.7777.../10 = (3 + 0.77777...)/10 = (3 + 7/(10^1 - 1))/10 = (3 + 7/9)/10 = (3 7/9)/10 = (34/9)/10 = 34/90
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According to Boyle's Law, if the temperature of a confined gas is held fixed, then the product of the pressure P and the volume
Bad White [126]

If <em>PV</em> = 800, then <em>P</em> can be written as a function of <em>V</em>,

<em>P(V)</em> = 800 / <em>V</em>

(a) The average rate of change of <em>P</em> as <em>V</em> increases from 200 to 250 in³ is then

(<em>P</em> (250) - <em>P</em> (200)) / (250 in³ - 200 in³) = (3.2 lb/in² - 4lb/in²) / (50 in³)

... = -0.016 (lb/in²)/in³

(Or -0.016 lb/in⁵, but I figure writing the rate as (units of pressure) per (unit volume) makes more sense.)

(b) We can also write <em>V</em> as a function of <em>P</em> :

<em>V(P)</em> = 800 / <em>P</em>

Take the derivative:

<em>V'(P)</em> = - 800 / <em>P</em>²

which immediately demonstrates that <em>V'(P)</em> ∝ 1 / <em>P</em>², as required. (The fish-looking symbol, ∝, means "is proportional to".)

If differentiating is supposed to be more involved, you can use the limit definition:

V'(P)=\displaystyle\lim_{h\to0}\frac{V(P+h)-V(P)}h

V'(P)=\displaystyle\lim_{h\to0}\frac{\frac{800}{P+h}-\frac{800}P}h

V'(P)=\displaystyle\lim_{h\to0}\frac{\frac{800P-800(P+h)}{P(P+h)}}h

V'(P)=\displaystyle800\lim_{h\to0}\frac{-\frac h{P(P+h)}}h

V'(P)=\displaystyle-800\lim_{h\to0}\frac1{P(P+h)}=-\dfrac{800}{P^2}

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3 years ago
The equation y = −3x + 7 represents the amount of water y left in a 7-gallon tub after x minutes. What is the slope of the line?
Novosadov [1.4K]
Y= -3x+7 = 2.3

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4 0
4 years ago
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Find the zeros of each function.
Anvisha [2.4K]

The zeros of the given functions are shown on the attached picture.

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Mary ran 2 1/0 laps. Ronald ran 1/3 laps around the lake as Mary did. How many laps around the lake did Ronald run?
Mila [183]

Answer:

you need to subtract

Step-by-step explanation:

just find the common denominator and i dont really know the rest i think you tube will help

3 0
3 years ago
Write down at least five number<br><br> pairs to solve the equation<br><br> (r – 2)(s +1) = 0.
Vika [28.1K]

Answer:

(r,s) => (2,-1);(2,0); (1,-1); (4,-1);(2,3)

Step-by-step explanation:

Given

(r - 2)(s +1) = 0

Required

Write at least 5 pair of numbers

(r - 2)(s +1) = 0

This can be split to be

r -2 = 0  --- (1)

or

s + 1 = 0 --- (2)

Solve for r in (1)

r = 2

<em>This means that for whatever value that s assumes, the equation will always be true when r = 2</em>

<em></em>

Solve for s in (2)

s = -1

<em>This means that for whatever value that r assumes, the equation will always be true when s = -1</em>

<em></em>

So, we can make use of the following pair

(r,s) => (2,-1);(2,0); (1,-1); (4,-1);(2,3)

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4 years ago
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