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Ray Of Light [21]
3 years ago
12

The greatest amount of rainfall is times as great as the least amount of rainfall

Mathematics
2 answers:
navik [9.2K]3 years ago
5 0

Answer:

16 times

Step-by-step explanation:

The question is basically asking :

<em>what number multiplied with the least amount would give me the greatest amount??</em>

<em />

<em>Let that number be x, so we can say:</em>

<em>least amount * x = greatest amount</em>

<em />

From the plot, we can see that least amount is 1/8 inches

The greatest amount is 2 inches

Now we can put it into the formula and find x:

1/8 * x = 2

x = 2/(1/8)

x = 16

Makovka662 [10]3 years ago
4 0

Answer:

16 times

Step-by-step explanation:

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valkas [14]
1/4, 8/15, 6/5
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Or you should know 1/4= .25 , 8/15 is around half or .5 and 6/5  the numerator is larger than the denominator, so it would be bigger than 1 <span />
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lutik1710 [3]

Answer:

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Step-by-step explanation:

diagonals bisect angles and opposite angles are congruent

therefore, ∠WXY ≅ ∠WZY

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Limit as x approaches infinity: 2x/(3x²+5)
Nonamiya [84]
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BUT, notice the denominator, for the same values of "x", the denominator becomes larg"er" than the numerator on every iteration, ever becoming larger and larger, and yielding a fraction whose denominator is larger than the numerator.

as the denominator increases faster, since as the lingo goes, "reaches the limit faster than the numerator", the fraction becomes ever smaller an smaller ever going towards 0.

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3 years ago
Find the solution of the differential equation dy/dt = ky, k a constant, that satisfies the given conditions. y(0) = 50, y(5) =
irga5000 [103]

Answer:  The required solution is y=50e^{0.1386t}.

Step-by-step explanation:

We are given to solve the following differential equation :

\dfrac{dy}{dt}=ky~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)

where k is a constant and the equation satisfies the conditions y(0) = 50, y(5) = 100.

From equation (i), we have

\dfrac{dy}{y}=kdt.

Integrating both sides, we get

\int\dfrac{dy}{y}=\int kdt\\\\\Rightarrow \log y=kt+c~~~~~~[\textup{c is a constant of integration}]\\\\\Rightarrow y=e^{kt+c}\\\\\Rightarrow y=ae^{kt}~~~~[\textup{where }a=e^c\textup{ is another constant}]

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and

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Thus, the required solution is y=50e^{0.1386t}.

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Answer:

SRT

Plzzzzz give me Brainliest!!!!!  

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