Alegbra 2 homework help?
1. solve the ineqaulity
|2x-3| < 5 2. Solve the inequality |2x - 5| > 1
3.A rectangle is 5ft longer than it is wide. The perimeter of the rectangle is the length of the rectangle
4.For a door to meet specifications at a carpentry shop, the width must be within 1/4in of the expected width of the door. The shop gets an order for doors that are 4ft wide. Write an inequality that expresses the range of widths for acceptable doors.
5.You have $15,000 available to invest in 2 stores, A and B. Write an inequality stating the restriction on A and B.
6. Solve for Z : y^2z + 7z = y
7. Solve for C: F=9/5C = 32
please help, and thank you if you do. most answered will definetely get the whole 10 points or whatever. and I have tried doing these..but..i just don’t get them, sorry. :\
Favorite Answer
2x – 3 < 5 x < 4 -(2x - 3) < 5 2x - 3 > -5
x > -1
so … -1 < x < 4 2. 2x - 5 > 1
x > 3
-(2x – 5) > 1
2x – 5 < -1 x < 2 so x > 3 or x < 2 3. I think you are missing something here ... l = w + 5 the perimeter can't be the length alone ... 4. 3.75 < x < 4.25 where x is the width of the door 5. A <= 15 000 and B <= 15 000 you can invest at most 15 000 in either store 6. z(y^2 + 7) = y z = y/ (y^2 + 7) 7. I assume you mean F = (9/5)C + 32 F - 32 = (9/5)C (5/9)F - 160/9 = C
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