All categories

2 years ago

Um...i am. confused, to say the least.

Mathematics

2 answers:

Jobisdone [24]2 years ago

8 0

Answer:

───▄▀▀▀▄▄▄▄▄▄▄▀▀▀▄───

───█▒▒░░░░░░░░░▒▒█───

────█░░█░░░░░█░░█──── this is Taco

─▄▄──█░░░▀█▀░░░█──▄▄─

█░░█─▀▄░░░░░░░▄▀─█░░█

░░░░░▐▀█▀▌░░░░▀█▄░░░

░░░░░▐█▄█▌░░░░░░▀█▄░░

░░░░░░▀▄▀░░░▄▄▄▄▄▀▀░░

░░░░▄▄▄██▀▀▀▀░░░░░░░

░░░█▀▄▄▄█░▀▀░░

░░░▌░▄▄▄▐▌▀▀▀░░ This is Bob

▄░▐░░░▄▄░█░▀▀ ░░ they have made an alliance

▀█▌░░░▄░▀█▀░▀ ░░ Copy and paste them so we can take over brainly.

░░░░░░░▀███▀█░▄░░

░░░░░░▐▌▀▄▀▄▀▐▄░░

░░░░░░▐▀░░░░░░▐▌░

░░░░░░█░░░░░░░░█░

Step-by-step explanation:

Nostrana [21]2 years ago

6 0

Answer:

Hello! Your answer would be, D) r = -4

Step-by-step explanation:

Hope I helped! Ask me anything if you have more questions! Have a nice mourning! Brainiest plz! Hope you make an 100%! -Amelia♥

You might be interested in

(7+5 + 4 . 13 - 2 Please simplify this expression with all steps thank you

FinnZ [79.3K]

Answer:

We apply the BODMAS rule

(7+5+4).(13-2)

(16).(11)

=176

Hope was useful

6 0

2 years ago

Read 2 more answers

Evaluate the following expressions if ×=2,y=3 and Z=4. a.4×+×-3over ×+10 b.2y-5yover y+5. c.yz-×. d.9+y+zover y+2

sammy [17]

Answer:

A= 7 11/12 or 7.916 or 95.12

B= 4 1/8 or 4.125 or 33/8

C= 12 4/5 or 12.8 or 64/5

6 0

2 years ago

Is this adjacent or vertical

vazorg [7]

Is there supposed to be an image? Adjacent is adjoining, so like touching each other. Vertical is "up". Horizontal is "side"

4 0

2 years ago

Read 2 more answers

1)A System of equations is shown below . What is the solution to the system of equations? 5x+2y=-15 2x-2y=-6

meriva

Answer:

x= -3 and y= 0

Step-by-step explanation:

5x+2y=-15

2x-2y=-6

7x =-21

x= -3

Putting value of x in equation 1

5(-3) +2y=-15

-15+2y= -15

2y= 0

y= 0

This can be solved with the help of matrices

In matrix form the above equations can be written in the form

$\left[\begin{array}{ccc}5&2\\2&-2\/\end{array}\right]$ $\left[\begin{array}{ccc}x\\y\\\end{array}\right]$ = $\left[\begin{array}{ccc}-15\\-6\\\end{array}\right]$

Let

$\left[\begin{array}{ccc}5&2\\2&-2\/\end{array}\right]$ = A $\left[\begin{array}{ccc}x\\y\\\end{array}\right]$ = X and $\left[\begin{array}{ccc}-15\\-6\\\end{array}\right]$ = B

Then AX= B

or X= A⁻¹ B

where A⁻¹= adj A/ ║A║ where mod A≠ 0

adj A= $\left[\begin{array}{ccc}-2&-2\\-2&5\/\end{array}\right]$

║A║= ( 5*-2- 2*2)= -10-4= -14≠0

X= A⁻¹ B

$\left[\begin{array}{ccc}x\\y\\\end{array}\right]$ =- 1/14 $\left[\begin{array}{ccc}-2&-2\\-2&5\/\end{array}\right]$ $\left[\begin{array}{ccc}-15\\-6\\\end{array}\right]$

$\left[\begin{array}{ccc}x\\y\\\end{array}\right]$ =- 1/14 $\left[\begin{array}{ccc}-2*-15&+ -2*-6\\-2*-15&+ 5*-6\\\end{array}\right]$

$\left[\begin{array}{ccc}x\\y\\\end{array}\right]$ =- 1/14 $\left[\begin{array}{ccc} 30&+12\\30&+-30\\\end{array}\right]$

$\left[\begin{array}{ccc}x\\y\\\end{array}\right]$ =- 1/14 $\left[\begin{array}{ccc}42\\0\\\end{array}\right]$

$\left[\begin{array}{ccc}x\\y\\\end{array}\right]$ = $\left[\begin{array}{ccc}-42/14\\0/-14\\\end{array}\right]$

$\left[\begin{array}{ccc}x\\y\\\end{array}\right]$ = $\left[\begin{array}{ccc}-3\\0\\\end{array}\right]$

From here x= -3 and y= 0

Solution Set = [(-3,0)]

3 0

2 years ago

Give a counterexample for the statement: 'All square roots are irrational numbers'. Explain your reasoning.

galben [10]

A counterexample for the statement 'All square roots are irrational numbers.' is the square root of 9. This is because when simplified, the answer is 3.

For a number to be irrational, it can't be expressed in fractional form. However, 3 can be expressed in fractional form as 3/1.

5 0

2 years ago