write the standard equation of a cirlce with the center -2,-4 and radius: 10
(x+2)^2 +(y+4)^2 = 10^2
okay what about the x and y intercepts for y= -I(abs.value) x+5I
for x- intercept put y=0 and for y-intercept put x=0 in y= -I(abs.value) x+5I
so would the equation just be y= x+5 since theres an absolute value mark on each side
dont consider the absolute
Write the slope-intercept form of the equation of the line through the given point and parallel to the given line. point : (3/5, 1/2) & line is 2x+5y=0
consider this equation first 2x+5y=0 can you write the slope intercept form of this equation??
isn't it y=-2/5x+0 ?
exactly
so what is the slope here???
-2/5, now what
ok you got the slope = -2/5 now two lines have the same slope when they are parallel thus the slope for the line you want to find is same as -2/5 now you have the slope = -2/5 and one point (3/5, 1/2) use point slope form now to get the equation
isn't the point slope form y-y1=m(x-x1) ???
yesssssss
k.one minute don't go anywhere I have more questions after this!
ok
right now im at y- 1/2= -2/5x + 6/25. I have to retrice1/2 to 6/25 but idk how to add those bc what could even be the common denominator to add it?
retrace mean ADD idk what that word was
ok its right now leave y alone what you get??
when I leave y alone I get : y= -2/5 + (6/25+1/2) how do I add those last two
by LCM
i know i have to find a common denominator but with 2 and 25 that's not possible
take the LCM of 25 and 2 first
thatd be 1 right?
bc 2 doesn't go into 25
no its 50
so would i multiply the numerator by 50? im confused at how this will be set up with a lcm of 50
ok i will draw the this step for you
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