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Lady_Fox [76]
2 years ago
7

PLEASE HELP! WILL MARK BRAINLIEST!

Mathematics
1 answer:
Kaylis [27]2 years ago
7 0

Answer:

320 meters

Step-by-step explanation:

x² + 200x = 166400

It's a square, area should be side²

x² + 2(x)(100) = 166400

x² + 2(x)(100) + 100² = 166400 + 100²

(x + 100)² = 176400

x + 100 = 420

Since these are sides, we just consider the positive square root

x + 100 = 420

x = 320m

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Find m <2 and M<4 kekekskskkskskskskskksksksksks
adoni [48]

Answer:

ok this is a good question

Step-by-step explanation:

i don't even know it

5 0
3 years ago
Given the domain {-4, 0, 5}, what is the range for the relation 12x 6y = 24? a. {2, 4, 9} b. {-4, 4, 14} c. {12, 4, -6} d. {-12,
xz_007 [3.2K]

The domain of the function 12x + 6y = 24 exists {-4, 0, 5}, then the range of the function exists {12, 4, -6}.

<h3>How to determine the range of a function?</h3>

Given: 12x + 6y = 24

Here x stands for the input and y stands for the output

Replacing y with f(x)

12x + 6f(x) = 24

6f(x) = 24 - 12x

f(x) = (24 - 12x)/6

Domain = {-4, 0, 5}

Put the elements of the domain, one by one, to estimate the range

f(-4) = (24 - 12((-4))/6

= (72)/6 = 12

f(0) = (24 - 12(0)/6

= (24)/6 = 4

f(5) = (24 - 12(5)/6

= (-36)/6 = -6

The range exists {12, 4, -6}

Therefore, the correct answer is option c. {12, 4, -6}.

To learn more about Range, Domain and functions refer to:

brainly.com/question/1942755

#SPJ4

7 0
1 year ago
g In R simulate a sample of size 20 from a normal distribution with mean µ = 50 and standard deviation σ = 6. Hint: Use rnorm(20
Illusion [34]

Answer:

> a<-rnorm(20,50,6)

> a

[1] 51.72213 53.09989 59.89221 32.44023 47.59386 33.59892 47.26718 55.61510 47.95505 48.19296 54.46905

[12] 45.78072 57.30045 57.91624 50.83297 52.61790 62.07713 53.75661 49.34651 53.01501

Then we can find the mean and the standard deviation with the following formulas:

> mean(a)

[1] 50.72451

> sqrt(var(a))

[1] 7.470221

Step-by-step explanation:

For this case first we need to create the sample of size 20 for the following distribution:

X\sim N(\mu = 50, \sigma =6)

And we can use the following code: rnorm(20,50,6) and we got this output:

> a<-rnorm(20,50,6)

> a

[1] 51.72213 53.09989 59.89221 32.44023 47.59386 33.59892 47.26718 55.61510 47.95505 48.19296 54.46905

[12] 45.78072 57.30045 57.91624 50.83297 52.61790 62.07713 53.75661 49.34651 53.01501

Then we can find the mean and the standard deviation with the following formulas:

> mean(a)

[1] 50.72451

> sqrt(var(a))

[1] 7.470221

5 0
2 years ago
I need to know if this is correct
Law Incorporation [45]
Yes its right
hope this helps


7 0
2 years ago
PLEASE HELP NEED ASAP WILL MARK THE BRAINLIEST
harkovskaia [24]

Answer:

idk

Step-by-step explanation:

idk

7 0
3 years ago
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