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ella [17]
3 years ago
14

(2a+b)^2 how do you do this

Mathematics
1 answer:
gulaghasi [49]3 years ago
3 0

•  Expand (2a + b)²:

(2a + b)²

= (2a + b) · (2a + b)


Multiply out the brackets by applying the distributive property of multiplication:

= (2a + b) · 2a + (2a + b) · b

= 2a · 2a + b · 2a + 2a · b + b · b

= 2²a² + 2ab + 2ab + b²


Now, group like terms together, and you get

= 2²a² + 4ab + b²

= 4a² + 4ab + b²    <———    expanded form  (this is the answer).


I hope this helps. =)


Tags:  <em>special product square of a sum algebra</em>

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I think its  

Step-by-step explanation:

I think its a because the IQR is the third quartile minus the first quartile (third-first=IQR)(the first is the middle number between the minimum and median and the third quartile is the middle number between the median and the maximum)so anyway it just makes sense that the answer would be B

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An underhand serve follows the same parabolic path but is hit from a height of 3 feet. How will this affect the focus and direct
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3 years ago
Solve the matrix equation for a, b, c, and d. [1 2] [a b] [6 5][3 4] [c d]= [19 8]
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Answer:

The answer is "\bold{\left[\begin{array}{cc}a&b\\c&d\end{array}\right] = \left[\begin{array}{cc}7&-2\\ -\frac{1}{2}&\frac{7}{2}\end{array}\right]}".

Step-by-step explanation:

\bold{\left[\begin{array}{cc}1&2\\3&4\end{array}\right] \left[\begin{array}{cc}a&b\\c&d\end{array}\right] = \left[\begin{array}{cc}6&5\\ 19&8\end{array}\right]}

Solve the L.H.S part:

\left[\begin{array}{cc}1&2\\3&4\end{array}\right] \left[\begin{array}{cc}a&b\\c&d\end{array}\right]\\\\\\\left[\begin{array}{cc}a+2c&b+2d\\3a+4c&3b+4d\end{array}\right]

After calculating the L.H.S part compare the value with R.H.S:

\left[\begin{array}{cc}a+2c&b+2d\\3a+4c&3b+4d\end{array}\right]= \left[\begin{array}{cc}6&5\\ 19&8\end{array}\right]} \\\\

\to a+2c =6....(i)\\\\\to b+2d =5....(ii)\\\\\to 3a+4c =19....(iii)\\\\\to 3b+4d = 8 ....(iv)\\\\

In equation (i) multiply by 3 and subtract by equation (iii):

\to 3a+6c=18\\\to 3a+4c=19\\\\\text{subtract}... \\\\\to 2c = -1\\\\\to  c= - \frac{1}{2}

put the value of c in equation (i):

\to a+ 2 (- \frac{1}{2})=6\\\\\to a- 2 \times \frac{1}{2}=6\\\\\to a- 1=6\\\\\to a =6 +1\\\\\to a = 7\\

In equation (ii) multiply by 3 then subtract by equation (iv):

\to 3b+6d=15\\\to 3b+4d=8\\\\\text{subtract...}\\\\\to 2d = 7\\\\\to d= \frac{7}{2}\\

put the value of d in equation (iv):

\to 3b+4 (\frac{7}{2})=8\\\\\to 3b+4 \times \frac{7}{2}=8\\\\\to 3b+14=8\\\\\to 3b =8-14\\\\\to 3b = -6\\\\\to b= \frac{-6}{3}\\\\\to b= -2

The final answer is "\bold{\left[\begin{array}{cc}a&b\\c&d\end{array}\right] = \left[\begin{array}{cc}7&-2\\ -\frac{1}{2}&\frac{7}{2}\end{array}\right]}".

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3 years ago
2 times what equals 36?
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3 years ago
(3x - 2)^5 =(3x - 2)^2
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Answer:

x=3/2,1

Step-by-step explanation:Given

(3x-2)ˆ5-(3x-2)ˆ2=0

(3x-2)ˆ3(3x-2)ˆ2-(3x-2)ˆ2=0

(3x-2)ˆ2{(3x-2)ˆ3-1}=0

(3x-2)ˆ2=0 Or (3x-2)ˆ3-1=0

3x-2=0 Or(3x-2)ˆ3=1

x=3/2 Or 3x-2=1, x=1

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3 years ago
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