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3 years ago

I need help for this one.

Mathematics

1 answer:

LuckyWell [14K]3 years ago

4 0

Answer:

Step-by-step explanation:

When you are looking for a pattern, look at one of the examples, makeup the pattern, then check it on the other examples.

Wrong answers may work on one, but not all the completed examples.

The right answer will work on all of the examples, then use that rule on the incomplete one.

So, I looked at the second one in the top row. The two bottom numbers, 5 + 13 makes 18. And 18÷3 (the top number, gives the center number) Check on other examples.

bottom + bottom=_

_÷top=center.

8+4=12, 12÷3 =4.

9+5=14, 14÷2=7

Both of these work.

So use the rule for the last one.

Add the bottom numbers, divide by the top number, fill it in the middle.

14+6=20

20÷4=5

Fill in 5.

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I will give brainliest but you need to explain how you received the answer so i can learn how to do it on my own/// thanks!

Readme [11.4K]

To solve this problem we must know that when any two lines intersect , a pair of opposite angles from the figure Will be equal
so that means that
$4x + 20 = 3x + 55$
we can subtract twenty from each side
$4x + (20 - 20) = 3x + (55 - 20)$
$4x = 3x + 35$
now we can subtract like terms
$4x - 3x = (3x - 3x) + 35$
so we can get the final answer as
$1x = 35$

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3 years ago

Which point is a solution to the inequality

julsineya [31]

HELLO! HOPE YOURE HAVING A GREAT DAY

your answer is C!

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3 years ago

The sides of a rectangle are in the ratio 5 : 4. If the perimeter of the rectangle is 360 cm , find its area

djverab [1.8K]

Answer:

Step-by-step explanation:

let, the length is z

thus the ratio will be 5z : 4z

perimeter = 2 (a + b)

360 = 2 ( 5z + 4z )

180 = 9z

z = 20 cm

length = 100 cm and width = 80cm

area = length * width

area = 100*80

area = 8000cm²

4 0

2 years ago

100 POINTS TO RIGHT AWNSER!!!!

makvit [3.9K]

If you're using the app, try seeing this answer through your browser: brainly.com/question/3000586

——————————

The answer is option D) r < 5 or r > – 1.

I'm going to graph each inequality below on a number line.

A) r > 5 or r > – 1.

$\large\begin{array}{cl} \mathsf{r>5}&\qquad\mathsf{\textsf{|||}\!\!\!\underset{-1}{\circ}\!\!\!\textsf{|||||}\!\!\underset{5}{\circ}\!\!\overset{*****~~~}{\textsf{|||}}\!\!\!\footnotesize\begin{array}{l}\blacktriangleright \end{array}\qquad \mathbb{R}}\\\\ \mathsf{r>-1}&\qquad\mathsf{\textsf{|||}\!\!\!\underset{-1}{\circ}\!\!\!\overset{********************~~~}{\textsf{|||||}\!\!\underset{5}{\bullet}\!\!\textsf{|||}}\!\!\!\footnotesize\begin{array}{l}\blacktriangleright \end{array}\qquad \mathbb{R}}\\\\\\ \mathsf{r>5~~or~~r>-1}&\qquad\mathsf{\textsf{|||}\!\!\!\underset{-1}{\circ}\!\!\!\overset{********************~~~}{\textsf{|||||}\!\!\underset{5}{\bullet}\!\!\textsf{|||}}\!\!\!\footnotesize\begin{array}{l}\blacktriangleright \end{array}\qquad \mathbb{R}} \end{array}$

The result is found just by joining those two intervals together. Actually that compound inequality only implies

r > – 1

which does not represent all real numbers.

—————

B) r > 5 or r < – 1.

$\large\begin{array}{cl} \mathsf{r>5}&\qquad\mathsf{\textsf{|||}\!\!\!\underset{-1}{\circ}\!\!\!\textsf{|||||}\!\!\underset{5}{\circ}\!\!\overset{*****~~~}{\textsf{|||}}\!\!\!\footnotesize\begin{array}{l}\blacktriangleright \end{array}\qquad \mathbb{R}}\\\\ \mathsf{r5~~or~~r$

Numbers between – 1 and 5 (including them) are not included in the union, so you don't have all real numbers represented there either.

—————

C) r < 5 or r < – 1.

$\large\begin{array}{cl} \mathsf{r$

Numbers that are greater or equal to 5 are not in the union. So it does not represent all real numbers.

—————

D) r < 5 or r > – 1.

$\large\begin{array}{cl} \mathsf{r-1}&\qquad\mathsf{\textsf{|||}\!\!\!\underset{-1}{\circ}\!\!\!\overset{********************~~~}{\textsf{|||||}\!\!\underset{5}{\bullet}\!\!\textsf{|||}}\!\!\!\footnotesize\begin{array}{l}\blacktriangleright \end{array}\qquad \mathbb{R}}\\\\\\ \mathsf{r-1}&\qquad\mathsf{\overset{~~~**************************~~~}{\textsf{|||}\!\!\!\underset{-1}{\bullet}\!\!\!\textsf{|||||}\!\!\underset{5}{\bullet}\!\!\textsf{|||}}\!\!\!\footnotesize\begin{array}{l}\blacktriangleright \end{array}\qquad \mathbb{R}} \end{array}$

Now all real numbers are included in the union. So this is the right choice.

Answer: option D) r < 5 or r > – 1.

I hope this helps. =)

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3 years ago

Use the box plots comparing the number of male puppies and number of female puppies attending doggy daycare each day for a month

dalvyx [7]

The IQR for the males' data is 25.
The difference between the median of the males' data and the female's data is 14.
The distribution of the males' data is skewed to the right and the median would be a better measure of the center. The distribution of the female's data is normal and the mean would be a better measure of the center.
A reason for the outlier is that the number of dogs needing care increased.

<h3>What is the interquartile range?</h3>

The interquartile range for the males data is the difference between the third quartile and the first quartile.

IQR = third quartile - first quartile

25 - 0 = 25

Median = 20 - 6 = 14

An outlier is a number that is way smaller or way larger than that of other numbers in a data set.

To learn more about outliers, please check: brainly.com/question/27197311

#SPJ1

7 0

2 years ago