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lesya [120]
3 years ago
12

Rewrite the expression using the distributive property and collecting like terms

Mathematics
2 answers:
timofeeve [1]3 years ago
5 0
12x

The fractions cancel out, so it’s just 9x+3x
Bumek [7]3 years ago
3 0
12x would be the answer
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6 0
2 years ago
Prove that<br>{(tanθ+sinθ)^2-(tanθ-sinθ)^2}^2 =16(tanθ+sinθ)(tanθ-sinθ)
USPshnik [31]

First, expand the terms inside the bracket you will get

(( \tan {}^{2} (x)  + 2 \tan(x)  \sin(x)  +  \sin {}^{2} (x)  - ( \tan {}^{2} (x)  - 2 \tan(x)  +  \sin {}^{2} (x) ) {}^{2}  = 16( \tan(x)  +  \sin(x) )( \tan(x)  -  \sin(x) )

( 4 \tan(x)  \sin(x) ) {}^{2}  = 16( \tan(x)  +  \sin(x) )( \tan(x)  -  \sin(x) )

16 \tan {}^{2} (x)  \sin {}^{2} (x)  = 16( \tan(x)  +  \sin(x) )( \tan(x)  -  \sin(x) )

16 \tan {}^{2} (x) (1 -  \cos {}^{2} (x) ) = 16 (\tan(x)  +  \sin(x) )( \tan(x)  -  \sin(x) )

16( \tan {}^{2} (x)  -   \frac{  \sin {}^{2} (x) \cos {}^{2} ( {x}^{} )  }{ \cos {}^{2} (x) }

16( \tan {}^{2} (x)  -  \sin {}^{2} (x) ) = 16( \tan(x)  +  \sin(x) )( \tan(x)  -  \sin(x) )

16( \tan(x)  +  \sin(x) )( \tan(x)  -  \sin(x)  = 16( \tan(x)  +  \sin(x) )( \tan(x)  -  \sin(x) )

5 0
2 years ago
Which ordered pair is a a solution of the equation ?
UNO [17]

Answer:

A

Step-by-step explanation:

when you plug in 2,11

11 = 3*2 +5 that is correct

13 = 3*3 +5 that is not correct

only 2,11 works

8 0
3 years ago
Read 2 more answers
A lumberjack can cut a log into 5 pieces in 20 minutes. How long does it take to cut a log of the same size and shape into 7 pie
Nitella [24]

Answer:

28minutes

Step-by-step explanation:

Use this simple trick, it can be used almost everywhere

Time taken pieces

20mins 5

? 7

An x shape is always multiply and straight line is divide

So (7 x 20) ÷5

If u still don't understand, just comment again

8 0
3 years ago
Pls pls help<br> Show work :)
Natasha2012 [34]

9514 1404 393

Answer:

  31.243 units

Step-by-step explanation:

The mnemonic SOH CAH TOA reminds you of the relationships between sides and angles in a right triangle. Using the attached figure, it is convenient to find the length of BE as an intermediate step in the solution.

  Sin = Opposite/Hypotenuse

  sin(30°) = BE/100

  BE = 100·sin(30°)

Then ...

  Tan = Opposite/Adjacent

  tan(58°) = BE/x

  x = BE/tan(58°) = 100·sin(30°)/tan(58°)

  x ≈ 31.243 . . . . units

_____

<em>Comment on the figure</em>

The intermediate problem in creating the figure was to locate point D. That was accomplished by locating point C on a line at an angle of 58° CCW from the horizontal, using point B as a center. Then D is the intersection of BC with the x-axis. BE is drawn perpendicular to the x-axis.

5 0
3 years ago
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