inequalities and gr 12 calculus?
the values of x that satisfy
|5-2x|<(or equal to)7 are?
a) x>(or equal to) -1
b) -6<x<1
c) -1<x<6
d) x<6
(all of the < signs also mean equal to)
question 2
if a line has a slope of 5 and passes through the points (-2,4) and (3,k) then k is equal to?
a) 9
b)29
c) 5
d) 21/5
e) none of the above?
question 3?
the equation of the lin taht is parallel to 5x-3y+2=0 has the same x-intercept as y=x^2-2x+1 is?
a) y= 3/5x-3/5
b) 5/3x-5/3
c) -5/3x+5/3
d) -3/5+3/5
e) none
THANKS SO MUCH IN ADVANCE. I REALLY NEED THESE ANSWERS.
Comments
✐Derivation✐
1. Create equations with negative and positive numbers and different inequality signs.
Positive:
5 - 2x ≤ 7
-2x ≤ 2
x ≥ -1
Negative:
5 - 2x ≥ -7
-2x ≥ -12
x ≤ 6
2. Recall that:
m = (y - y1)/(x - x1)
5 = (4 - k)/(-2 - 3)
5 = (4 - k)/-5
-25 = 4 - k
-k = -29
k = 29
3. Recall that:
y = mx + b
m is the slope.
b is y-intercept.
Parallel - the slopes are congruent to each other.
It's easier to make the equations into slope-intercept form.
5x - 3y + 2 = 0
-3y = -5x - 2
y = (-5x - 2)/-3
y = (5/3)x + (2/3)
That's first equation. Take out a slope with 5/3 and substitute it to the new equation.
y = (5/3)x + b
Now, the x-intercept. Make y = 0.
Let's find the x-intercept of the second equation.
y = x² - 2x + 1
0 = x² - 2x + 1
0 = (x - 1)²
x = 1
So the x-intercept is 1. The new equation should have the x-intercept of 1 too.
0 = (5/3)(1) + b
0 = 5/3 + b
b = -5/3
y = (5/3)x - (5/3)