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9966 [12]
2 years ago
14

+{-x2y+[-(x2y-2xy2+y2)+(-xy2-3y2+x2y)]-(10y2-x2y)}

Mathematics
1 answer:
Kay [80]2 years ago
6 0

Answer:

3xy² -  14y²      

Step-by-step explanation:

I hope that this is the problem

   - x²y + [ - (x²y - 2xy² + y²) + (xy² - 3y² + x²y)] - (10y² - x²y)

= - x²y + [ - x²y + 2xy² - y² + xy² - 3y² + x²y] - 10y² + x²y

Now combine like terms in the [  ].

= - x²y  + [ -x²y + x²y + 2xy² + xy² - y² - 3y² ] - 10y² + x²y

= - x²y  + [ 0 + 3xy² - 4y²] - 10y² + x²y

= - x²y + 3xy² - 4y² -10y² + x²y                      Now combine like terms

=   (-x²y + x²y) + 3xy² + (-4y² - 10y²)

=       0 + 3xy² - 14y²

=  3xy² -  14y²       or    y²(3x - 14)

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Step-by-step explanation:

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Jerry paid $99.30 for 6 shirts.how much did jerry pay for each shirt?
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Step-by-step explanation:

3 0
2 years ago
Delaney would like to make a 10 lb nut mixture that is 60% peanuts and 40% almonds. She has several pounds of peanuts and severa
scoundrel [369]

Answer:

a) p + m = 10 and \frac{p + 0.2m}{0.8m} = \frac{60}{40}

b) The mixing amount is 5 lb peanut and 5 lb mixture.

Step-by-step explanation:

Given that p pounds of peanuts and m pounds of the 80% almond and 20% peanut mixture are used to make 10 pounds of 60% peanuts and 40% of almonds mixture.

Now, we can write that p + m = 10 ........ (1)  

Now, m pounds of 80% almond and 20% peanut mixture contain 0.2m pounds of peanuts and 0.8m pounds of almonds.

Now, from the condition given it can be written that  

\frac{p + 0.2m}{0.8m} = \frac{60}{40} .......... (2)

⇒ \frac{p + 0.2m}{0.8m} = \frac{3}{2}

⇒ 2p + 0.4m = 2.4m

⇒ 2p = 2m

⇒ p = m  

Now from equation (1) we get p = m = 5 pounds.

a) Therefore, equations (1) and (2) are the system of equations that models the situation.

b) The mixing amount is 5 lb peanut and 5 lb mixture. (Answer)

3 0
3 years ago
Read 2 more answers
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