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bagirrra123 [75]
3 years ago
7

What equation (Ex: Y=mx+b) would have an intersection with x=y on the origin (0,0)? PLEASE HELP!!

Mathematics
1 answer:
vivado [14]3 years ago
5 0
(2, -1) I think, I could be wrong.
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What is the equivalent of 6x-8y?​
liberstina [14]
Plug y=-5 into the equation
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Mary leaves her house to take a walk. The graph shows the distance, d, in feet from
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D. 35, since that would be the longest amount. of time walking away from her house, she should be the farthest amount of feet away from her house
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Please help! I will not need help with the graph at the bottom just the equation
gladu [14]

isolate y

move y to the right

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solution is

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6 0
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So you final answer is:  81.25 hours.

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8 0
3 years ago
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
6 0
3 years ago
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