y < - 8 or y > 4
inequalities of the form | x | > a always have solutions of the form
x < - a or x > a
we have to solve
y + 2 < - 6 or y + 2 > 6
y + 2 < - 6 ( subtract 2 from both sides )
y < - 8
or
y + 2 > 6 ( subtract 2 from both sides )
y > 4
these can be combined using interval notation
y ∈ (- ∞, - 8 ) ∪ (4, ∞ )
As a check
substitute chosen values of x from each interval
y = - 10 : | - 10 + 2 | = | - 8 | = 8 > 6 this is true
y = 12 : | 12 + 2 | = | 14 | = 14 > 6 which is also true
Answer:
solo suma todos los valores x+4(3)+(x+x)
en el segundo x+5(2)+x+3(2)
en el tercero x-3+(2x+5)+(2x+5)
y en el cuarto x(3)+(x+4)
Answer:
D. 2
Step-by-step explanation:
Common denominator is already presented which is 4.
-3/4 and 3/4 cancel each other out.
you can also convert to decimal form -3/4 + 2 3/4 is the same as:
-0.75 + 2.75 = 2.
