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GuDViN [60]
2 years ago
6

Add: -5g^2+(-9g^2-8)

Mathematics
1 answer:
Amiraneli [1.4K]2 years ago
3 0

Answer:

\boxed{-14g^2 - 8}

Step-by-step explanation:

-5g^2 + (-9g^2 - 8)

= - 5g^2 - 9g^2 - 8

= -14g^2 - 8

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WHAT IS X³-27 SIMPLIFIED
Eduardwww [97]

Answer:

<u>It</u><u> </u><u>is</u><u> </u><u>(</u><u>x</u><u> </u><u>-</u><u> </u><u>3</u><u>)</u><u>³</u><u> </u><u>-</u><u> </u><u>9</u><u>x</u><u>(</u><u>3</u><u> </u><u>-</u><u> </u><u>x</u><u>)</u>

Step-by-step explanation:

Express 27 in terms of cubes, 27 = 3³:

=  {x}^{3}  -  {3}^{3}

From trinomial expansion:

{(x - y)}^{3}  = (x - y)(x - y)(x - y) \\

open first two brackets to get a quadratic equation:

{(x - y)}^{3}  = ( {x}^{2}  - 2xy +  {y}^{2} )(x - y)

expand further:

{(x - y)}^{3}  =  {x}^{3}   - y {x}^{2}  - 2y {x}^{2}  + 2x {y}^{2}  + x {y}^{2}  -  {y}^{3}  \\  {(x - y)}^{3}  =  {x}^{3}  -  {y}^{3}  + 3x {y}^{2}  - 3y {x}^{2}  \\  {(x - y)}^{3}  =  {x}^{3}  -  {y}^{3}  + 3xy(y - x) \\  \\ { \boxed{( {x}^{3} -  {y}^{3} ) =  {(x - y)}^{3}   - 3xy(y - x)}}

take y to be 3, then substitute:

( {x}^{3}  - 3^3) =  {(x - 3)}^{3}  - 9x(3 - x)

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2 years ago
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2 years ago
Please help. My son and I don't get it
vfiekz [6]
So you have 3 and you want to get rid of 2 2/5. 3-2 is 1  1-2/5 is equal to three fifths, which is the answer. 
6 0
2 years ago
Read 2 more answers
Paul was thinking of a number. Paul adds 10, then divides by 2 to get an answer of -13. What was the original number?
frozen [14]

Answer:

I believe its -36

Step-by-step explanation:

-36+10=-26

-26/2= -13

3 0
2 years ago
Read 2 more answers
The difference of two numbers is 16 the greater number is five less than four times a small number find the two numbers
Elis [28]

Answer:

The numbers are 23 and 7  

Step-by-step explanation:

1. Set up the equations

   Let x = the greater number

  and y = the smaller number. Then

(1) x - y = 16

       4y = 4 times the smaller number and

  4y - 5 = 5 less than 4 times the smaller number. Then

(2)     x = 4y - 5

You have a system of two equations:

\begin{cases}(1) & x - y = 16\\(2) & x = 4y - 5\end{cases}

2. Solve the equations

\begin{array}{lrcll}(3) & 4y - 5 - y & = & 16 &\text{Substituted (2) into (1)}\\& 3y - 5 & = & 16 &\text{Simplified}\\ & 3y & = & 21 &\text{Added 5 to each side}\\(4) &y & = & \mathbf{7} &\text{Divided each side by 3} \\& x - 7 & = & 16 & \text{Substituted (4) into (1)}\\& x& = & \mathbf{23} & \text{Added 7 to each side}\\\end{array}

The larger number is 23; the smaller number is 7.

3. Check

\begin{array}{rclcrcl}23 - 7& = & 16 & \qquad &23&=&4(7) - 5\\16 & = & 16 & \qquad &23& = &28 - 5\\&&& \qquad &23& = &23\\\end{array}

OK.

3 0
3 years ago
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