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Artemon [7]
2 years ago
7

Rewrite the expression...see attached picture....

Mathematics
2 answers:
USPshnik [31]2 years ago
7 0
Yes you should use the first choice
MakcuM [25]2 years ago
7 0
For this case we must take into account the following properties of exponents:
 1) Same base, the exponents are added
 2) The addition and subtraction of polynomials can only be done between terms of the same degree.
 3) r (x) is the residue
 4) q (x) is the quotient
 5) b (x) is the divisor
 Answer:
 
See attached image.

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statuscvo [17]
A = l x w
A = (x+2) ( 3x)
A = 3x^2 + 6x

answer
Area of a rectangle = 3x^2 + 6x
3 0
3 years ago
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Dave ran 3.52 miles in the same time
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Subtract 3.52 - 2.44= 1.08 miles
8 0
3 years ago
If tanA+sinA=m and tanA-sinA=n.prove that m^2-n^2=4√mn
mash [69]

Let a=\tan A and b=\sin A. Then

m^2-n^2=(a+b)^2-(a-b)^2=(a^2+2ab+b^2)-(a^2-2ab+b^2)=4ab

\implies m^2-n^2=4\tan A\sin A

and

mn=(a+b)(a-b)=a^2-b^2

\implies4\sqrt{mn}=4\sqrt{\tan^2A-\sin^2A}

The expression under the square root can be rewritten as

\tan^2A-\sin^2A=\dfrac{\sin^2A}{\cos^2A}-\sin^2A=\sin^2A\left(\dfrac1{\cos^2A}-1\right)=\sin^2A(\sec^2A-1)

Recall that

\sin^2A+\cos^2A=1\implies\tan^2A+1=\sec^2A

so that

\tan^2A-\sin^2A=\sin^2A\tan^2A

and assuming \sin A>0 and \tan A>0, we end up with

4\sqrt{\tan^2A-\sin^2A}=4\tan A\sin A

so that

m^2-n^2=4\sqrt{mn}

as required.

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Answer:

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Step-by-step explanation:

Hope this helps! Sorry if its wrong

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2 years ago
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