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It is: y = x+3 whereas 1 is the slope and 3 is the y intercept

Q: What is the equation of the line whose intercept is 3 and slop is 1?

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It is the intercept.

-10

The line will be going 'uphill' from left to right

4x+5y = 20 5y = -4x+20 y = -4/5x+4 So the slope is -4/5 or as -0.8 and the y intercept is 4

It is a straight line equation that needs to be rearranged into slope-intercept form as folows:- 7x+2y = 5 Subtract 7x from both sides: 2y = -7x+5 Divide all terms by 2: y = -3.5x+2.5 which is now in slop-intercept form

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It is the intercept.

If you mean the slope of the line with equation y = x then it is 1 and there is no y intercept

y = mx + b where m is the slope and b is the y-intercept

-10

slop intercept form is y=mx+b where m is the slope. We have to find b. To do so we have to evaluate the problem. We see that the y-intercept is 3 which means the coordinate pair (x,y)=(0,3). Plug (0,3) into the question. it should like this 3=m(0)+b. now solve. anything times 0 is 0 so the equation is now 3=b. now we have m, which is -2 and b which is 3. plug it into the equation and the slop intercpet form is y=-2x+3 =D

The line will be going 'uphill' from left to right

4x - 5y = 0 standard form of slope intercept line equation is y = mx + c m = slop and c = intercept rewrite the 4x - 5y = 0 in standard form y = 4/3 x compare it with standard form, and slope = 4/5

y = -4x is a line with a slop of -4 and a y-intercept of 0 (it passes through the point (0,0).

y=4/3x-4

4x+5y = 20 5y = -4x+20 y = -4/5x+4 So the slope is -4/5 or as -0.8 and the y intercept is 4

It is a straight line equation that needs to be rearranged into slope-intercept form as folows:- 7x+2y = 5 Subtract 7x from both sides: 2y = -7x+5 Divide all terms by 2: y = -3.5x+2.5 which is now in slop-intercept form

It is tilted up as you trace it going left to right. If you write the equation for the line in y = mx + b form, m being positive means that a positive step in x results in a positive step in y. If the intercept b happened to be zero, then the line would lie exclusively in the I and III quadrants.