Okay I really don't get this :
Part 1: Solve the inequality -5x + 8 x - 4 and show all of your work. You can use <= to represent the less than or equal to symbol.
Part 2: Use complete sentences to describe the graph of the solution from Part 1. Be sure to include the type of circle and which direction the shading will be.
ALSO Which of the following is not an equivalent form of the compound inequality
x - 1 < -11 or x + 12 > 14
and this last one I still don't get Which of the following ordered pairs is a solution to the inequality y -2x + 8?
please help with whichever you can, Thank you so much
Part 1: Solve the inequality -5x + 8 x - 4 and show all of your work. You can use <= to represent the less than or equal to symbol.
Part 2: Use complete sentences to describe the graph of the solution from Part 1. Be sure to include the type of circle and which direction the shading will be.
ALSO Which of the following is not an equivalent form of the compound inequality
x - 1 < -11 or x + 12 > 14
and this last one I still don't get Which of the following ordered pairs is a solution to the inequality y -2x + 8?
please help with whichever you can, Thank you so much
-
1)
-5x + 8 ≤ x - 4
4 + 8 ≤ x + 5x
12 ≤ 6x
2 ≤ x
it will be closed circle shaded to the right since we are greater than or equal to 2.
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2)
x - 1 < -11. . . . . or. . . . x + 12 > 14
x < -11 + 1. . . . . or. . . . x > 14 - 12
x < -10. . . . . or. . . . x > 2
(-∞ , -10) U (2 , ∞)
a.A number line with an open circle on -10, an open circle on 2, and shading in between.
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3)
d.(0, -4)
y ≤ -2x + 8 ---> (0 , -4)
-4 ≤ -2 * 0 + 8
-4 ≤ 8 <--- true
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4)
c.(-1, 0)
y ≥ -4x - 2 ----> (-1 , 0)
0 ≥ -4 * -1 - 2
0 ≥ 4 - 2
0 ≥ 2 <--- not true
=======
free to e-mail if have a question
-5x + 8 ≤ x - 4
4 + 8 ≤ x + 5x
12 ≤ 6x
2 ≤ x
it will be closed circle shaded to the right since we are greater than or equal to 2.
-----------
2)
x - 1 < -11. . . . . or. . . . x + 12 > 14
x < -11 + 1. . . . . or. . . . x > 14 - 12
x < -10. . . . . or. . . . x > 2
(-∞ , -10) U (2 , ∞)
a.A number line with an open circle on -10, an open circle on 2, and shading in between.
----------------
3)
d.(0, -4)
y ≤ -2x + 8 ---> (0 , -4)
-4 ≤ -2 * 0 + 8
-4 ≤ 8 <--- true
----------
4)
c.(-1, 0)
y ≥ -4x - 2 ----> (-1 , 0)
0 ≥ -4 * -1 - 2
0 ≥ 4 - 2
0 ≥ 2 <--- not true
=======
free to e-mail if have a question