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Vesna [10]
3 years ago
14

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<br>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

Thus, the general solution is ye^(2x) - 3e^(2x) = C or (y - 3)e^(2x) = C.

7 0
3 years ago
Numbers 2-4 Will choose brainliest
Sever21 [200]

Answer:

2. 9, 8+9

3. 17-8=9, 17, 8, 9

4. 13-7=6

Step-by-step explanation:

6 0
3 years ago
What is exspanded notation for 3.6 and 4
Ede4ka [16]
It is shown as a sum of each digit multiplied by its matching place value<span> (units, tens, hundreds, etc.) For example: 4,265 = </span>4<span> x 1,000 + 2 x 100 + 6 x 10 + 5 x 1.</span>
8 0
3 years ago
Emma is finding 7/2 x 2/5
Eva8 [605]

Answer:

14/10

OR

7/5

Step-by-step explanation:

(7*2)/(2*5) = 14/10

(14/2) / (10/2) = 7/5

7 0
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

What length of gestation, rounded to the nearest whole number, represents the 18th percentile?

This is X when Z has a pvalue of 0.18. So X when Z = -0.915.

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

-0.915 = \frac{X - 284}{5.5}

X - 284 = -0.915*5.5

X = 279

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

8 0
3 years ago
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