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SCORPION-xisa [38]
3 years ago
8

determine the most precise name of the quadrilateral ABCD from the information given. for #13 & #14

Mathematics
2 answers:
Vika [28.1K]3 years ago
8 0
Hello,
----------------------
Answer ⇒ <u>14</u>
----------------------
snow_tiger [21]3 years ago
4 0
Answer:14\\\\because\ in\ 13\ \overline{AE}=\overline{CA}\ and\ \overline{BE}=\overline{DE}\ is\ a\ rhombus.
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Scott has seven dogs if four of them are Doberman how many are not Doberman A) N-4=7 , B ) 7=n+4 , C) n-7=4 D) 4+7=n
Bond [772]
The answer is obviously 3 but for the equation to use would be N=7-4 but looking at the choices I'd go with B
6 0
3 years ago
Identify the level of measurement (nominal, ordinal, or interval-ratio) of each of the following variables: (1) How satisfied a
Allisa [31]

Solution :

Nominal variable

A nominal variable is defined as a variable which is used to \text{nam}e \text{  or label or categorize some particular attributes }  which are being measured.

An ordinal variable is the one in which the order matters, but the difference between any two orders does not matter.

In interval ratio variable is defined as the variable where the difference between any two values is meaningful.

The level of measurement  for each of the following are :

1) Variables that are categorized in categories so that it is ordinal data.

2) Data scaled with the two categories her or his,  so it is a nominal data.

3) Number of votes categorized in the intervals so it is Interval type data.

4) nominal data.

3 0
3 years ago
phineas has more jolly ranchers than alan. if alan has 10 jolly ranchers how many do they have in total
maxonik [38]

Answer:

We cannot know.

Step-by-step explanation:

It only mentions that Phineas has more jolly ranchers than Alan. Since Alan has 10 jolly ranchers, the number of jolly ranchers Phineas could have is 11, 12, 13, 14... and so on. Phineas could have 1 billion jolly ranchers.

5 0
2 years ago
????????????????????????????
Anna [14]
If you type in both the coordinates it’ll give you the answer on google. That’s how I did this.
3 0
3 years ago
Read 2 more answers
#24, i am getting <img src="https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%260%260%26%5Cfrac%7B18%7D%7B7%7D%20
sveticcg [70]

It looks like you're talking about row-reducing an augmented matrix to solve the system of equations. Your answer is almost correct. The last row should read 0, 0, 1, 2/7.

The given system translates to

\left[ \begin{array}{ccc|c} 2 & -3 & 1 & 2 \\ 1 & -1 & 2 & 2 \\ 1 & 2 & -3 & 4 \end{array} \right]

Eliminate x from the last two rows by combining -2 (row 2) and row 1, and -2 (row 3) and row 1; that is,

(2x - 3y + z) - 2 (x - y + 2z) = 2 - 2 (2)

2x - 3y + z - 2x + 2y - 4z = 2 - 4

-y - 3z = -2

and

(2x - 3y + z) - 2 (x + 2y - 3z) = 2 - 2 (4)

2x - 3y + z - 2x - 4y + 6z = 2 - 8

-7y + 7z = -6

In augmented matrix form, this step yields

\left[ \begin{array}{ccc|c} 2 & -3 & 1 & 2 \\ 0 & -1 & -3 & -2 \\ 0 & -7 & 7 & -6 \end{array} \right]

I'll omit these details in the remaining steps.

Eliminate y from the last row by combining -7 (row 2) and row 3 :

\left[ \begin{array}{ccc|c} 2 & -3 & 1 & 2 \\ 0 & -1 & -3 & -2 \\ 0 & 0 & 28 & 8 \end{array} \right]

Multiply the last row by 1/28 :

\left[ \begin{array}{ccc|c} 2 & -3 & 1 & 2 \\ 0 & -1 & -3 & -2 \\ 0 & 0 & 1 & 2/7 \end{array} \right]

Eliminate z from the second row by combining 3 (row 3) and row 2 :

\left[ \begin{array}{ccc|c} 2 & -3 & 1 & 2 \\ 0 & -1 & 0 & -8/7 \\ 0 & 0 & 1 & 2/7 \end{array} \right]

Multiply the second row by -1 :

\left[ \begin{array}{ccc|c} 2 & -3 & 1 & 2 \\ 0 & 1 & 0 & 8/7 \\ 0 & 0 & 1 & 2/7 \end{array} \right]

Eliminate y and z from the first row by combining 3 (row 2), -1 (row 3), and row 1 :

\left[ \begin{array}{ccc|c} 2 & 0 & 0 & 36/7 \\ 0 & 1 & 0 & 8/7 \\ 0 & 0 & 1 & 2/7 \end{array} \right]

Multiply the first row by 1/2 :

\left[ \begin{array}{ccc|c} 1 & 0 & 0 & 18/7 \\ 0 & 1 & 0 & 8/7 \\ 0 & 0 & 1 & 2/7 \end{array} \right]

8 0
3 years ago
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