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2 years ago

What property is shown in the example below?

Mathematics

2 answers:

Pavlova-9 [17]2 years ago

7 0

Answer:

Associative

Step-by-step explanation:

noname [10]2 years ago

5 0

D is the correct answer.

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Find the distance between the points ( – 18,4) and ( – 18, – 20).

worty [1.4K]

Answer:

(-18,4) is 16 units below (-18,20). hope this helps

Step-by-step explanation:

- Zombie

6 0

2 years ago

Given the function ƒ(x ) = 3x + 1, evaluate ƒ(a + 1).

Setler79 [48]

Answer:

$\Huge \boxed{3a+4}$

Step-by-step explanation:

The function is given.

$f(x)=3x+1$

To find : $f(a+1)$

The input for the function f(x) is (a + 1).

Replace x with (a + 1).

$f(a+1)=3(a+1)+1$

Expand brackets.

$f(a+1)=3a+3+1$

Simplifying.

$f(a+1)=3a+4$

3 0

3 years ago

Mathematically verify the outlier(s) in the data set using the 1.5 rule.

statuscvo [17]

Given:

The data values are:

7, 8, 11, 13, 14, 14, 14, 15, 16, 16, 18, 19

To find:

The outliers of the given data set.

Solution:

We have,

7, 8, 11, 13, 14, 14, 14, 15, 16, 16, 18, 19

Divide the data set in two equal parts.

(7, 8, 11, 13, 14, 14), (14, 15, 16, 16, 18, 19)

Divide each parenthesis in two equal parts.

(7, 8, 11), (13, 14, 14), (14, 15, 16), (16, 18, 19)

Now,

$Q_1=\dfrac{11+13}{2}$

$Q_1=\dfrac{24}{2}$

$Q_1=12$

And

$Q_3=\dfrac{16+16}{2}$

$Q_3=\dfrac{32}{2}$

$Q_3=16$

The interquartile range is:

$IQR=Q_3-Q_1$

$IQR=16-12$

$IQR=4$

The data values lies outside the interval $[Q_1-1.5IQR,Q_3+1.5IQR]$ are known as outliers.

$[Q_1-1.5IQR,Q_3+1.5IQR]=[12-1.5(4),16+1.5(4)]$

$[Q_1-1.5IQR,Q_3+1.5IQR]=[12-6,16+6]$

$[Q_1-1.5IQR,Q_3+1.5IQR]=[6,22]$

All the data values lie in the interval [6,22]. So, there are no outliers.

Hence, the correct option is 4.

3 0

3 years ago

Please help me out with this!! BRAINLIEST AVAILABLE!!

lapo4ka [179]

Answer:

xy = 1

k = 79

Step-by-step explanation:

Question One

The first and third frames look to me to be the same. I'll treat them that way.

y = x^2 Equate y = x^2 to the result of 2y + 6 = 2x + 6

2y + 6 = 2(x + 3) Remove the brackets

2y + 6 = 2x + 6 Subtract 6 from both sides

2y = 2x Divide by 2

y = x

Now solve these two equations.

so x^2 = x

x > 0

1 solution is x = 0 from which y = 0. This won't work. x must be greater than 0. So the other is

x(x) = x Divide both sides by x

x = 1

y = x^2 Put x = 1 into x^2

y = 1^2 Solve

y = 1

The second solution is

(1,1)

xy = 1*1

xy = 1

Answer: A

Question Two

square root(k + 2) - x = 0

Subtract x from both sides

sqrt(k + 2) = x Square both sides

k + 2 = x^2 Let x = 9

k + 2 = 9^2 Square 9

k + 2 = 81

k = 81 - 2

k = 79

4 0

3 years ago

A^3-a^2-2a+8 factorize

Nostrana [21]

Answer:

factors-----(a+2)(a^2+4-3a)

4 0

2 years ago