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3 years ago

What is the scale factor of two regular pentagons that have areas of 50cm square and 162cm square.

Mathematics

1 answer:

4vir4ik [10]3 years ago

5 0

Answer:

(m/n)

Step-by-step explanation:

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horsena [70]

Start with

$-2x-4 < 3x+21$

Add 2x to both sides:

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3 years ago

Read 2 more answers

Sin(x+pi/4)-sin(x-pi/4)=1 solve the equation

evablogger [386]

Sin(α+β)=sin(α)cos(β)+cos(α)sin(β)
sin(α-β)=sin(α)cos(β)-cos(α)sin(β)

$sin(x+ \frac{ \pi }{4} )=sin(x)cos\frac{ \pi }{4} +cos(x)sin\frac{ \pi }{4} \\ sin(x- \frac{ \pi }{4} )=sin(x)cos\frac{ \pi }{4} -cos(x)sin\frac{ \pi }{4} \\ \\ \\sin(x+ \frac{ \pi }{4} )-sin(x- \frac{ \pi }{4} ) =1 \\ sin(x)cos\frac{ \pi }{4} +cos(x)sin\frac{ \pi }{4} -(sin(x)cos\frac{ \pi }{4} -cos(x)sin\frac{ \pi }{4} )=1 \\ sin(x)cos\frac{ \pi }{4} +cos(x)sin\frac{ \pi }{4} -sin(x)cos\frac{ \pi }{4} +cos(x)sin\frac{ \pi }{4} =1 \\ cos(x)sin\frac{ \pi }{4} +cos(x)sin\frac{ \pi }{4} =1 \\$
$2cos(x)sin\frac{ \pi }{4} =1 \\ sin\frac{ \pi }{4} = \frac{ \sqrt{2} }{2} \\ 2cos(x) \frac{ \sqrt{2} }{2} =1 \\ 2 \cdot \frac{ \sqrt{2} }{2}cos(x) =1 \\ \sqrt{2} cos(x)=1 \\$
$cos(x)= \frac{1}{ \sqrt{2} } \\ x=\pm arccos \frac{1}{ \sqrt{2} }+2 \pi k , k \in Z \\ x=\pm \frac{ \pi }{4} +2 \pi k , k \in Z$

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3 years ago