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3 years ago

Please help!

Mathematics

1 answer:

scZoUnD [109]3 years ago

8 0

Answer: the first left equation is matching the description.

Step-by-step explanation

Since we know A= s2 ( exponent 2) and sis the side :

A=s2

A= 99sin2

s=square(99) sin

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Find the general solution of dy/dx + 2y =6

Marianna [84]

Answer:

ye^(2x) - 3e^(2x) = C or (y - 3)e^(2x) = C

Step-by-step explanation:

dy/dx + 2y = 6

This is a first order linear ODE.

p(x) = 2

Integrating factor: I(x) = e^(∫ 2 dx) = e^(2x)

e^(2x)(dy/dx + 2y) = 6e^(2x)

e^(2x) dy + 2ye^(2x) dx = 6e^(2x) dx

Integrate both sides of the equation.

ye^(2x) = 3e^(2x) + C

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3 years ago

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Answer:

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Step-by-step explanation:

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3 years ago

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Ede4ka [16]

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3 years ago

Emma is finding 7/2 x 2/5

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Answer:

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2 years ago

Read 2 more answers

In a certain breed of cattle, the length of gestation has a mean of 284 days and a standard deviation is 5.5 days. What length o

Vaselesa [24]

Answer:

A gestation length of 279 days represents the 18th percentile.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In a certain breed of cattle, the length of gestation has a mean of 284 days and a standard deviation is 5.5 days.

This means that $\mu = 284, \sigma = 5.5$