1)Find the range of the relation [(-1,4)},(2,5),(3,5)}.Then determine whether the relation is a funtion.
4,5,5 is the range
4,5 is a function
2)Find f(-1),if f(x)= x^2-6x/x+2
(-1)^2-6(-1)/-1+2
1+6/-1+2
7/1 =7
3)Find f(a),if f(t)= 2t^2-t-2.
2(a)^2-a-2
2a^2-a-2
4)Which equation is linear?
a.x=-2 (my answer)
b.y=3x^2+1
c.y=square root of x-2
d.y^2=1/2x+3
5)Write -3y = -1+5x in standard form.
-5x-3y=1
6)Find the x and y intercept of the graph of 4x-2y=8
4(0)-2y=8
0-2y=8
8/-2y= y=-4
4x-2(0)=8
4x=8
8/4x= x=2 so the answer is -4;2
7)Find the slope that passes through (2,4)and(-7,8).
8-4/-7-2= 4/-9
8)What is the slope of the line x= -2?
0
9)What is the slope of a line that is parallel to the graph of 2x-3y=6?
-2/3
10)The graph of the line through(2,3) that is perpendicular to the line with equation x= -1 also goes through which point?
(-2,3)
11)Write an equation in slope intercept form for the line that has a slope of 3 and passes through (-1,2).
y=3x+5
12)Write an equation inslope intercept form for the line that passes through (-1,-2)and (3,-7).
y= -5/4x-13/4
13)Write an equation for the line that passes through(0,5)and is parallel to the line whose equation is 4x-y=3.
y=4x+5
14)Evaluate f(3/4) if f(x)= [[1-2x]]
no clue
15)Identify the range of y=[x]-4
no clue
16)Which point satisfies the inequality y<_[-x+3]?
no clue
SORRY for all the problems at once I just thought it would be better for 1 post so please don't give me a hard time and thanks for all the help.
1. ok
2. and 3. write your solution in proper form, using brackets
e.g. for f(x)= x^2-6x/x+2
f(-1) = ((-1)^2 - 6(-1))/(-1+2)
=7/1
=7
4. ok
5. It is considered better form to have your equation start with a positive x term. You can always do that by simply changing all the signs.
so 5x + 3y = -1
6. ok
7. use brackets, and a proper statement
slope = (8-4)/(-7-2)
= - 4/9
8. NO. x=-1 is a vertical line, the slope of a vertical line is undefined. The slope of a horizontal line is 0
9. No, the slope is 2/3.
Change it into the form y=mx + b and you will see.
10. yes
11. yes
12. yes (you could have checked this by substituting the given points in your answer. They should both verify.
13. yes
14. are those absolute value signs???
If so then, f(3/4) = │1 - 2(3/4)│
=│-1/2│
= 1/2
15. y=[x]-4
The absolute value of x has its smallest value when x=0. So y ≥ -4
16. pick any point for x, eg. let x=5
then y ≤ │-2│
y ≤ 2
so pick a y value < 2, how about y=1
so the point (5,1) satisfies your inequation.
14. are those absolute value signs???
If so then, f(3/4) = „ 1 - 2(3/4)„
=„ -1/2„
= 1/2
no.they're just 2 brackets on each side. 1/2 is not a choice
on #16 I don't have that as a choice. I worked it out, so would (-2,4) work?
yes for 16, there would be an infinite number of choices.
for 15 it would then by -1/2
lol 14 is a wierd problem. My choices are 0,-2,-1,1 I think its called a special function problem
sorry i am too busy
5 answers
1.- (13,-5) (13,7)
2.- (5,-7/5) (-4,2/5)
4x− y = −8