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Nuetrik [128]
3 years ago
15

A rectangle’s length is three times its width, w. Its area is 243 square units. Which equation can be used to find the width of

the rectangle?
w2 = 3(243)
3w2 = 243
4w2 = 243
3w2 = 3(243)
Mathematics
2 answers:
kykrilka [37]3 years ago
8 0

Answer:

the second option because it's three times the width so it would be time the W

Step-by-step explanation:

Sauron [17]3 years ago
3 0

Answer:

the guy above me is correct

Step-by-step explanation:

just did the test got a 100%

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lorasvet [3.4K]
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3 years ago
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mrs_skeptik [129]

Answer:

g(4x) = 192x^3

Step-by-step explanation:

For this problem, f(x) is irrelevant since we are simply are dealing with g(x).  We will simply replace the value of x in g(x) with 4x.  So let's do that.

g(x) = 3x^3

g(4x) = 3(4x)^3

g(4x) = 3(4^3)(x^3)

g(4x) = 3(64)(x^3)

g(4x) = 192x^3

Hence, g(4x) is 192x^3.

Cheers.

4 0
3 years ago
If A+B+C=<img src="https://tex.z-dn.net/?f=%5Cpi" id="TexFormula1" title="\pi" alt="\pi" align="absmiddle" class="latex-formula"
seraphim [82]

Answer:

a + b + c = \pi \\  =  > c=  \pi - a - b \\  <  =  >  \tan(c)  =  \tan(\pi - a - b)  =  -\tan(a + b)

Step-by-step explanation:

we have:

\tan(a)  +  \tan(b)  +  \tan(c)  \\  =  \tan(a)  +  \tan(b)  -  \tan(a + b)  \\  =  \tan( a)  +  \tan(b)  -  \frac{ \tan(a) +  \tan(b)  }{1 -  \tan(a)  \tan(b) }  \\  =  \frac{ ( \tan(a) +  \tan(b)  ) \tan(a) \tan(b)  }{ \tan(a) \tan(b)  - 1 } (1)

we also have:

\tan(a)  \tan(b)  \tan(c)  \\  =  -  \tan(a)  \tan(b)  \tan(a + b)  \\  =  \frac{ -(\tan( a  )   + \tan(b) ) \tan(a)  \tan(b) }{1 -  \tan(a)  \tan(b) }  \\  =  \frac{( \tan(a)  +  \tan(b)) \tan(a)   \tan(b) }{ \tan(a) \tan(b)  - 1 } (2)

from (1)(2) => proven

5 0
3 years ago
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iris [78.8K]

Answer:

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Step-by-step explanation:

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