which answer is a solution to the inequality 4+|t+2|<11
A. t<5 and t>9
B. t>5 or t<-9
C.-5
first thing you may want to do is isolate the absolute value function
how do i do that?
take 4 off both sides
ok
so it will look like this |t+2|<11??
you took 4 from th eleft side, you have to do that to both sides to not change the equation
|t+2|<7 like this?
yes, The absolute value of something, here t+2, can be thought of as a distance from zero, and distance is a positive value weather you are at a - value or + value, the distance from 0 is positive
you have to think a bit differently weather the absolute value is greater or less than some value. Here it is less than, the distance is less than 11 , starting at a point, you can move +11 or -11 and the distance from the start is still +11, that is the general idea of absolute value
I am sorry, remember you took the 4 off, it is 7 not 11. so if it is a less than, you are stuck between moving +7 and moving - 7 \[-7 <t+2<+7\]
so t in that case is on the interval of -9 to +5 , -9<t<5 |dw:1447214479741:dw|
on the other hand if you simplify do isolate the absolute value, and you have that it is "Greater than " some value... then the distance has to be "at least some number"
so the answer is C?
example, just change the sign to greater in the last prob abs(t+2) > 11 here you get two seperated intervals, the distance is at least a certain amount, solve for both t + 2 =11 or t + 2 =-11
right, C for the other sign, you get an OR instead of a compound AND inequality
t>9 or t<-13
a split interval... here is a reference i found/ http://www.purplemath.com/modules/absineq.htm
Alright thank you :))
welcome
before you go can you check my answer on my last question please?
Select the graph that represents the solution of the open sentence. 3|p| – 5 < 7
k
isolate the absolute value first add 5 both sides divide by 3 both sides abs(p) < 4
less than means the distance is at most 4 in either direction from where you start, zero -4<p<4 good
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