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levacccp [35]
3 years ago
7

What is 2.1 as a fraction?

Mathematics
1 answer:
german3 years ago
7 0
2 1/10 because the 1 in 2.1 is in the tenth place
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5k^3-8-4k^2+5k^2-2+3k^3
lara31 [8.8K]
<h3>I'll teach you how to solve 5k^3-8-4k^2+5k^2-2+3k^3</h3>

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5k^3-8-4k^2+5k^2-2+3k^3

Group like terms:

5k^3+3k^3-4k^2+5k^2-8-2

Add similar elements:

5k^3+3k^3+k^2-8-2

Add similar elements:

8k^3+k^2-8-2

Subtract the numbers:

8k^3+k^2-10

Your Answer Is 8k^3+k^2-10

Plz mark me as brainliest :)

7 0
3 years ago
WILL MAKE BRAINLIEST!!!!!!!<br> PLEASE INCLUDE FULL WORKING!!!!
aleksandrvk [35]
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4 0
3 years ago
Do 2(x, y) → (3,5).<br> The point (x,y) is
Elan Coil [88]

Answer:

(1.5, 2.5)

Step-by-step explanation:

(x,y) is being multiplied by 2 to get (3,5). We need to divide it by 2 to find x and y.

3/2= 1.5

5/2=2.5

Thus, the answer is (1.5, 2.5). Hope this helped!

8 0
3 years ago
Please answer this question for brainliest!
kkurt [141]

The original side length is given as 3.00

Multiply the original length by the scale factor:

3.00 x 0.75 = 2.25

The answer is B. 2.25 cm



3 0
3 years ago
Find the solution to this system of equations <br> 3x+2y+3z=3 4x-5y+7z=1 2x+3y-2z=6 <br> x=? Y=? Z=?
Ede4ka [16]

Answer:

The solution is x=2,\ y=0,\ z=-1.

Step-by-step explanation:

You are given the system of three equations:

\left\{\begin{array}{l}3x+2y+3z=3\\4x-5y+7z=1\\2x+3y-2z=6\end{array}\right.

Multiply the first equation by 4, the second equation by 3 and subtract them. Then multiply the third equation by 2 and subtract it from the second equation:

\left\{\begin{array}{l}3x+2y+3z=3\\4(3x+2y+3z)-3(4x-5y+7z)=4\cdot 3-3\cdot 1\\4x-5y+7z-2(2x+3y-2z)=1-2\cdot 6\end{array}\right.\Rightarrow \\\\\left\{\begin{array}{l}3x+2y+3z=3\\12x+8y+12z-12x+15y-21z=12-3\\4x-5y+7z-4x-6y+4z=1-12\end{array}\right.

So,

\left\{\begin{array}{rl}3x+2y+3z=3\\23y-9z=9\\-11y+11z=-11\end{array}\right.\Rightarrow \left\{\begin{array}{l}3x+2y+3z=3\\23y-9z=9\\y-z=1\end{array}\right.

Multiply the third equation by 23 and subtract it from the second equation:

\left\{\begin{array}{rl}3x+2y+3z=3\\23y-9z=9\\23y-9z-23(y-z)=9-23\cdot 1\end{array}\right.\Rightarrow \left\{\begin{array}{rl}3x+2y+3z=3\\23y-9z=9\\23y-9z-23y+23z=9-23 \end{array}\right.

Hence,

\left\{\begin{array}{rl}3x+2y+3z=3\\23y-9z=9\\14z=-14 \end{array}\right.\Rightarrow z=-1

Substitute it into the second equation:

23y-9\cdot (-1)=9\Rightarrow 23y+9=9\\ \\23y=0\\ \\y=0

Substitute them into the first equation:

3x+2\cdot 0+3\cdot (-1)=3\Rightarrow 3x-3=3\\ \\3x=6\\ \\x=2

The solution is x=2,\ y=0,\ z=-1.

3 0
3 years ago
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