Ask your own question, for FREE!
Mathematics 18 Online
OpenStudy (anonymous):

Solving Inequalities by Multiplying or Dividing Questions?

OpenStudy (anonymous):

OpenStudy (anonymous):

OpenStudy (anonymous):

OpenStudy (anonymous):

OpenStudy (rajat97):

for question 3, the first in this list , option 3 is correct choice if we solve the equation , we get 0=0 on both the sides so it indicates that the equation is true for all values of x and has infinite solutions

OpenStudy (rajat97):

for question 4, the secondin this list, option 2nd is correct as the equality indicates that x is less than or has its value equal to -2 and the dot at-2 indicates that it will be included in the interval from which x belongs

OpenStudy (rajat97):

for question 6, you just need to follow basic mathematical operations but the only difference is that there is an inequality sign instead of the equal to sign so by solving this, you'll get the answer as option 2 as the correct option

OpenStudy (anonymous):

Thank you very much

OpenStudy (rajat97):

for question 7, you have to do the same that is you have to carry out the simple mathematical operation that is division you can directly write p<96/12 if the number before p(coefficient of p) would be -VE, then you would have changed the inequality for example -12p<96 12p>96 so p>8 so this is the method for multiplication and division and it's my pleasure to help you

OpenStudy (rajat97):

question 8 uses the method that i have illustrated in the previous post so it's answer will be g>90

OpenStudy (anonymous):

For question 7 wouldn't it be p<8

OpenStudy (rajat97):

you sure?

OpenStudy (rajat97):

question 9 says that result of 6 subtractracted from a number is AT LEAST 2 so it can be two and greater than that so by doing this , you can find out the answer and it comes out to be the 2nd option hope this helps you

OpenStudy (anonymous):

yes, well because all my answer choices have "<" sign so i was just wondering. and again thank you

OpenStudy (rajat97):

it was pleasure to help you

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!