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3 years ago

Determine the median of the following numbers: 8,11,11,12,13,15,16,16,18,21

Mathematics

1 answer:

lubasha [3.4K]3 years ago

6 0

Answer:

Step-by-step explanation:

Since 13 and 15 are the two middle numbers left after crossing out all the others, you add the two together and divide by 2 to get the average.

13+15=28

28/2= 14

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What is the value of X?

Naddika [18.5K]

Answer:

x= 80

Step-by-step explanation:

70 +30=180

sum of angles is triangle is 180 degree

sp 180-100= 80

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3 years ago

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15points+brainliest!! please help i really need this done

aev [14]

Answer:

2 units right, 5 units down

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3 years ago

Given that

valentina_108 [34]

Answer: x= -3.25

Step-by-step explanation:

R=4x+6y 11=4x+6(4) 11=4x+ 24 11-24=4x -13=4x -13/4=x x=-3.25

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3 years ago

{(-3, -5), (-1, 1), (0, 4), (1, 7)} list the domain amd range.

podryga [215]

The domain are the numbers in the X coordinates. So it would be -3, -1, 0, 1

The range is the numbers that are the Y coordinates. -5, 1, 4, 7

8 0

3 years ago

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What is the following sum? Assume x greater-than-or-equal-to 0 and y greater-than-or-equal-to 0

Free_Kalibri [48]

Answer:

$\sqrt{x^2y^3}+2\sqrt{x^3y^4}+xy\sqrt{y}=2xy^2\sqrt{x}+2xy\sqrt{y}$

Hence, ption B is true.

Step-by-step explanation:

Given the expression

$\sqrt{x^2y^3}+2\sqrt{x^3y^4}+xy\sqrt{y}$

solving the expression

$\sqrt{x^2y^3}+2\sqrt{x^3y^4}+xy\sqrt{y}$

$\sqrt{x^2y^3}=xy\sqrt{y}$

$2\sqrt{x^3y^4}=2xy^2\sqrt{x}$

so the expression becomes

$\:\sqrt{x^2y^3}+2\sqrt{x^3y^4}+xy\sqrt{y}=xy\sqrt{y}+2xy^2\sqrt{x}+xy\sqrt{y}$

Group like terms

$=2xy^2\sqrt{x}+xy\sqrt{y}+xy\sqrt{y}$

Add similar elements

$=2xy^2\sqrt{x}+2xy\sqrt{y}$

Therefore, we conclude that:

$\sqrt{x^2y^3}+2\sqrt{x^3y^4}+xy\sqrt{y}=2xy^2\sqrt{x}+2xy\sqrt{y}$

Hence, option B is true.

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3 years ago

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