Answer:
the answer is 23 and 24
Step-by-step explanation:
I just did the middle school test and that was the correct answer.
Answer:
Step-by-step explanation:
If a cubes diagonal measures you can solve for its sidelength by creating a triangle. Visualize the diagonal as the hypotenuse of our triangle inside the cube. Then, one of the legs would be the side of the cube itself, and the other leg would be the diagonal of the bottom side of the cube. Knowing this is a cube, we can call each side s. If each side is s, then the diagonal of the bottom face is ( this is a special triangle property, triangle.) Then we know our triangle is a right triangle, so we can use the Pythagorean theorem to solve for s.
Solve this equation for s.
Since we now found the side length of the cube, we can find its volume by cubing the side length:
The compound inequalities for each number line is 18: −1≤x≤4 19: −3<x≤2 20: x<0 or (1≤x≤3) 21: (−4<x≤−6) or (−3<x≤0) 22: (3≤x<1) or (6<x<7) 23: (−3≤x≤−4) or (0<x<2).
let's break down the compound inequalities for each situation step by step:
18:
−1≤x≤4 (closed on both ends)
The notation ≤ means "less than or equal to," so
−1≤x indicates that
x can be equal to or greater than -1.
Likewise,
x≤4 means that
x can be equal to or less than 4.
Combining both inequalities, we have
−1≤x≤4, indicating that
x can take any value between -1 and 4, including -1 and 4.
19:
−3<x≤2 (open on the left end, closed on the right end)
The notation < means "less than," so −3<x indicates that
x must be greater than -3 but not equal to -3.
x≤2 means that
x can be equal to or less than 2.
Combining both inequalities, we have
−3<x≤2, indicating that x can take any value greater than -3 and up to and including 2. It's open at -3 but closed at 2.
20: x<0 or (1≤x≤3) (open on the left end, closed on the right end, with a gap between 0 and 1)
x<0 indicates that x must be less than 0.
(1≤x≤3) indicates that x can be equal to or greater than 1 and equal to or less than 3.
Combining these two inequalities with "or," we have
x<0 or
1≤x≤3, indicating that x can be less than 0 or take any value between 1 and 3, including 1 and 3.
21:
(−4<x≤−6) or
(−3<x≤0) (open on both ends with a gap between -3 and -4)
(−4<x≤−6) indicates that x must be greater than -4 but not equal to -4, and it can be equal to or less than -6.
(−3<x≤0) indicates that x must be greater than -3 but not equal to -3, and it can be equal to or less than 0.
Combining these two inequalities with "or," we have
(−4<x≤−6) or
(−3<x≤0), indicating that
x can take any value greater than -4 and up to -6, or greater than -3 and up to 0.
22:
(3≤x<1) or
(6<x<7) (closed on the left end, open on the right end, with a gap between 1 and 6)
(3≤x<1) indicates that x can be equal to or greater than 3 and less than 1.
(6<x<7) indicates that x must be greater than 6 and less than 7.
Combining these two inequalities with "or," we have
(3≤x<1) or (6<x<7), indicating that x can take any value equal to or greater than 3 and less than 1, or greater than 6 and less than 7.
23:
(−3≤x≤−4) or
(0<x<2) (closed on the left end, open on the right end, with a gap between -4 and 0)
(−3≤x≤−4) indicates that x can be equal to or greater than -3 and equal to or less than -4.
(0<x<2) indicates that
x must be greater than 0 but less than 2.
Combining these two inequalities with "or," we have
(−3≤x≤−4) or
(0<x<2), indicating that x can take any value equal to or greater than -3 and equal to or less than -4, or greater than 0 and less than 2.
These compound inequalities describe the ranges of values for x in each given situation on the number line.
To know more about inequality:
#SPJ3
Answer:18-1 closed right 5
19-3 open right5
20 0 open left 1 3 closed right 1
21 -4 open left 2 -3 open right 3
22 3 close left 2 6 open right 1
23-3 open left 1 0 open right 2
Step-by-step explanation:
Answer:
I believe it would be the third option. x=20.
Step-by-step explanation:
the 2 over the x cancels out the square root over the 20.