I Have A Few Questions Relating To Inequalities.

Help with math inequalities!?!?!?

Is the inequality 2x-y���6 the same as y≥-2x-6 ?
............... No; I'm afraid not .................

Just remember that when you move anything to either side of the inequality, you just change the sign. And its all right to multiply and divide by positive numbers, but every time that you multiply or divide by a negative number, you change the direction of the inequality .... And be sure to try not to multiply or divide by a variable .... it might be negative.

Now let's take your inequality:

2x - y ≥ 6
2x ≥ 6 +y ........ o.k.
2x - y - 6 ≥ 0 ... o.k
y ≥ 6 - 2x ........ o.k.
x ≥ 3 + y/2 ...... o.k.
2 ≥ (6 +y) /x .... no
- 2x + y ≥ - 6 ... no
2x/y - 1 ≥ 6/y .. no

Help with inequalities?

1. You have two answers due to the fact you dont know what the part in the absolute value bars equal (so the answer is either)
a. 5x - 7 > 2
X > 9/5

b. 5x - 7 < -2
X < 1

So the answer is x > 9/5 or x < 1


2. Not sure. Factor? But with the calculator you can find the limits. If you have the calculator I think you, you do this:
a. Press the y = button
b. Input the problem
c. Press the Graph button
d. Look at it. You have to see where the graph has limits. Look on google to find lessons. I think I may just confuse you if I try to help.

3. Square both sides (square root cancels). Giving you:
5x - 4 = 16
5x = 20
x = 4

Test x = 4
(5(4) - 4) = 16
16 = 16

4 works! Yay!

Hope it helps

Help on inequalities ???

1. x² + ax + 11/x + 1 > 3

=> x³ + ax² - 2x + 11 > 0

But no cubic, whatever the value of a, can ever be greater than any given value for all x. So the question is meaningless. Are you sure you copied it right?

2. Again there appears to be no minimum, because there is nothing to stop y being as large-negative as you like (adjusting the value of a large positive number x to make the equation fit). So again, a meaningless question.

Why would I go on?...

I have a few questions relating to EARLY PREGNANCY.?

I didn't find out I was pregnant until I was six weeks pregnant but I sort of knew I was pregnant around five weeks. The only symptom I had was a missed period but I was irregular so I wasn't sure if it was pregnancy or just the irregular crap. I had a feeling in the pit of my stomach that I was pregnant. Around nine weeks is when I started to get more symptoms like tender breasts, sleeping more than usual and morning sickness.

About getting pregnant on your period: It's possible — but highly unlikely. You'd have to have a very short menstrual cycle, which is the time from the first day of one period to the first day of the next period, or a tendency to have long periods. This would bring the time of ovulation closer to the time when you would start bleeding.

Conception occurs when an egg and sperm meet in a fallopian tube. Sometime during the middle of your menstrual cycle, most likely between the 12th and 16th days, an egg reaches maturity in one of the two ovaries. The ovary releases the egg into the abdomen, where it's quickly sucked up by the tulip-shaped opening of the nearest fallopian tube.

Another site:
Can I get pregnant during my period?

Pregnancy can occur from intercourse that takes place during a period. This is because sperm can live in the body for up to five days, and if a woman ovulates soon after her period, then conception could take place from intercourse that occurred during her period.

INEQUALITY RELATING TO SHOPPING(MUST HAVE A VARIABLE)?

You can get so many pairs of shoes at $29.99 each, but you only have $80:

29.99x <= 80

Inequalities?

I'm not an expert on this, but I played with it for fun. a^b + b^a > 1 ln(a^b + b^a) > ln(1) ln(a^b) + ln(b^a) > 0 b*ln(a) + a*ln(b) > 0 b*ln(b) > -a*ln(a) (expr 1): b*ln(b)/a*ln(a) > -1 Note that: ln(x) = 0 for x = 1 ln(x) > 0 for x > 1 ln(x) < 0 for x >= 0 < 1 ln(x) approaches -infinity when x approaches 0 ln(x) is imaginary for x < 0 So if a and b are > 1 then expr 1 is true And if a and b are > 0 < 1 then expr 1 is true. And if a and b = 0 then expr 1 is true. And if a and b = 1 then expr 1 is true And if b = 1 and a > 0 expr 1 is true And if a or b < 0 then expr 1 is false if b > 1 and a >=0 < 1 it gets tricky: if b*ln(b) >= 1 and -1 < a*ln(a) < 0 then expr 1 will be false If b >= 1.763 and a < 1 then expr 1 is false So: if b > 1.763 and a > 1 then expr 1 is true It seems as though there are infinite possibilities and several different ranges of values that will work. I guess I'm not sure what "belonging to (0,1)" means exactly. Update: If a and b are > 0 < 1 then expr 1 is true because the LHS of expr 1 will always be positive and hence greater than RHS of expr 1 (-1).