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Sergio [31]
2 years ago
5

HELP MEEE???? L'LL GIVE U A BRAINLIST???

Mathematics
1 answer:
Katarina [22]2 years ago
3 0
Okay, don’t freak out. It’s more simple than it seems.
To find the area of this rectangle you use length x length
Let’s look for the length first
Use the bottom left and right points
(-2.25,-1) and (2.25,-1)
The length is a horizontal line so we use the x-coordinates. The distance from -2.25 to 2.25 is 4.5
Now let’s find the height. Let’s use the top left and bottom left coordinates.
(-2.25,4) and (-2.25,-1)
The height is a vertical line. The y-coordinates. Their difference is 5.
So 4.5 x 5 = 22.5
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Find the mean and range.<br><br> 26 19 23 39 31 34 23 25
sveticcg [70]
Range: largest number minus the smallest number
39 - 19 = 20
your range is 20

Mean: add all of the numbers and divide by the amount of numbers there are.

26+19+23+39+31+34+23+25= 220
220/8 = 27.5

your mean is 27.5

hope this helps
7 0
3 years ago
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Please solve <br> -(6x+7)+8=19
JulijaS [17]
-6x-7+8=19 and then add your like terms so it’s -6x+1=19 and then subtract one on both sides so it’s -6x=18 so x=-3
3 0
3 years ago
453.2 divided by 14 (Round to the nearest tenth)
Bond [772]

Answer:

{453.2 \div 14} \\  =  \frac{453.2}{14}  \\  \\  = { \boxed{32.4}}

3 0
2 years ago
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Use matrices and elementary row to solve the following system:
LiRa [457]

I assume the first equation is supposed to be

5x-3y+2z=13

and not

5x-3x+2x=4x=13

As an augmented matrix, this system is given by

\left[\begin{array}{ccc|c}5&-3&2&13\\2&-1&-3&1\\4&-2&4&12\end{array}\right]

Multiply through row 3 by 1/2:

\left[\begin{array}{ccc|c}5&-3&2&13\\2&-1&-3&1\\2&-1&2&6\end{array}\right]

Add -1(row 2) to row 3:

\left[\begin{array}{ccc|c}5&-3&2&13\\2&-1&-3&1\\0&0&5&5\end{array}\right]

Multiply through row 3 by 1/5:

\left[\begin{array}{ccc|c}5&-3&2&13\\2&-1&-3&1\\0&0&1&1\end{array}\right]

Add -2(row 3) to row 1, and add 3(row 3) to row 2:

\left[\begin{array}{ccc|c}5&-3&0&11\\2&-1&0&4\\0&0&1&1\end{array}\right]

Add -3(row 2) to row 1:

\left[\begin{array}{ccc|c}-1&0&0&-1\\2&-1&0&4\\0&0&1&1\end{array}\right]

Multiply through row 1 by -1:

\left[\begin{array}{ccc|c}1&0&0&1\\2&-1&0&4\\0&0&1&1\end{array}\right]

Add -2(row 1) to row 2:

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

Multipy through row 2 by -1:

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

The solution to the system is then

\boxed{x=1,y=-2,z=1}

5 0
3 years ago
What is the degree of the polynomial 3xy^2z−2z^5+2/3y^2z^4 ?
pav-90 [236]

Answer:

6 is the answer


4 0
3 years ago
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