The only linear equation is choice d
Answer:
Step-by-step explanation:
3 red, 1 blue, and 4 yellow...thats a total of 8 marbles
so a red was drawn and not replaced....means we now have 7 marbles.
P(yellow) : since we still have 4 yellow and there are a total of 7 marbles, the probability of drawing a yellow is : 4/7 <==
Answer:
1. y=1/4x+2
2. y=-3/2x +1
3. y = -3/2x -2
4. y= -3x -49
5. y = -2x +8
Step-by-step explanation:
1. Use slope intercept, y=mx+b where m =1/4 and b=2. y=1/4x+2
2. Use the point-slope form to write the equation, then simplify and convert into the slope intercept form.
(y-7)=-3/2(x--4)
y-7=-3/2(x+4)
y-7=-3/2x-6
y = -3/2x -6 +7
y=-3/2x +1
3. To write the equation of a line we must have a slope and a point. To find the slope we use the slope formula and substitute (x,y) points in it as shown below:
Now that we have the slope, plug in the slope and choose one point to plug into the point slope formula. Use the point-slope form to write the equation, then simplify and convert into the slope intercept form.
(y--5)=-3/2(x-2)
y+5=-3/2(x-2)
y+5=-3/2x + 3
y = -3/2x -2
4. Use the point-slope form to write the equation, then simplify and convert into the slope intercept form. The slope is -3 since parallel lines have the same slope.
(y--4)=-3(x-5)
y+4=-3 (x+15)
y+4=-3x -45
y= -3x -49
5. Use the point-slope form to write the equation, then simplify and convert into the slope intercept form. The slope is -2 since perpendicular lines have the negative reciprocal slopes. So 1/2 becomes -2.
(y--6)=-2(x-7)
y+6=-2 (x-7)
y+6 = -2x + 14
y = -2x +8
Answer:
C) One vertical but no horizontal
Step-by-step explanation:
Use implicit differentiation to find dy/dx.
xy² = 2 + xy
x (2y dy/dx) + y² = x dy/dx + y
(2xy − x) dy/dx = y − y²
dy/dx = (y − y²) / (2xy − x)
When the tangent lines are horizontal, dy/dx = 0.
0 = (y − y²) / (2xy − x)
0 = y − y²
0 = y (1 − y)
y = 0 or 1
When the tangent lines are vertical, dy/dx is undefined, so the denominator is 0.
0 = 2xy − x
0 = x (2y − 1)
x = 0, y = 1/2
There are two possible points where there can be a horizontal tangent line, and two possible points where there can be a vertical tangent line. Plug all back into the original equation to see which are actual points.
If y = 0:
x(0)² = 2 + x(0)
0 = 2
No solution.
If y = 1:
x(1)² = 2 + x(1)
x = 2 + x
0 = 2
No solution.
If x = 0:
(0)y² = 2 + (0)y
0 = 2
No solution.
If y = 1/2:
x(1/2)² = 2 + x(1/2)
x/4 = 2 + x/2
-x/4 = 2
x = -8
Therefore, there is one vertical line and no horizontal lines.