simplify:(a+2b) ^3-(a-2b) ^3​

Answer:

Step-by-step explanation:

3.(2y+1)×(y2−2y3+3y)

=2y(y2−2y3+3y)+1(y2−2y3+3y)

=2y^3−4y^4+6y^2+y^2−2y^3+3y

=−4y^4+2y^3−2y^3+6y^2+y^2+3y

=−4y^4+7y^2+3y

Always use distributive property for solving these kinds of questions.

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*note: when multiplying two same variables,always add the powers.......

for eg: x³ × x⁴ = x⁷*

Step : multiply 2x with each term in the other polynomial....

2x × x²-2x -1 = (2x × x²) + (2x × -2x) + (2x × -1)

= 2x3 - 4x² - 2x

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Step 1 : multiply x⁵ with each term in the other polynomial....

x⁵ × x³+2x²+2 = (x⁵ × x³) + (x⁵ × 2x²) + (x⁵ × 2)

=. x⁸ + 2x⁷ + 2x⁵ ----> (a)

Step 2: multiply -1 with each term in the other polynomial.......

-1 × x³+2x²+2 = (-1 × x³) + (-1 × 2x²) + (-1 × 2)

= -x³-2x²-2 ------->(b)

Step 3: now add the two products....

(x⁸ + 2x⁷ + 2x⁵) + (-x³-2x²-2)

=x⁸+2x⁷+2x⁵-x³-2x²-2

2y × y² - 2y³ + 3y = (2y × y²) + (2y × 2y³) + (2y × 3y)

= 2y³ + 4y⁴ + 6y²

1 × y² - 2y³ + 3y = (1 × y²) + (1 × 2y³) + ( × 3y)

= y² - 2y³ + 3y

(2y³ + 4y⁴ + 6y²) + (y² - 2y³ + 3y)

= 2y³ - 2y³ + 4y⁴ + 6y² -y² +3y

= 4y⁴ + 6y² -y² +3y

Answer:

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Explanation:

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Answer:

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Answer:

14 correct answer answer answer answer

Two identical cars means , they have equal mass

K.E.=

2

1

​

mv

2

Let m be the mass of each car

Kinetic energy of car A =  

2

1

​

m×(15)

2

=

2

225

​

m

Kinetic energy of car B =  

2

1

​

m×(30)

2

=450m

K.E.ofB

K.E.ofA

​

=

2×450m

225m

​

=

4

1

​

=1:4

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Answer:

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