Need help on this one thank u

A store sells mugs. Each mug cost the same amount.

Rom4ik [11]

Answer:

nobody won't be able to answer this without some more details.

4 0

3 years ago

Read 2 more answers

A girl throws a 2 kg rock with a net force of 6 N. What is the acceleration of the rock?

krek1111 [17]

Answer:
acceleration = 3 m/s²

Explanation:
From Newton's second law:
Force = mass * acceleration
We are given that:
Force = 6 Newton
mass = 2 kg

Substitute with the givens in the above equation to get the acceleration as follows:
Force = mass * acceleration
6 = 2 * acceleration
acceleration = 6 / 2 = 3 m/s²

Hope this helps :)

8 0

3 years ago

The greatest integer function, shown below, is defined so that it produces the greatest integer __ or equal to x. f(x) = [ [x] ]

scoundrel [369]

Less than or equal to x.

6 0

3 years ago

Read 2 more answers

Which expression shows the result of applying the distributive property to 3(2xC – 6)?

Nat2105 [25]

Answer:

Step-by-step explanation:

4 0

3 years ago

I NEED HELP ON THESE 2 PROBLEMS. pls good anawers or bad rating and report

vekshin1

Answer:

1.a. skew

1.b. parallel

1.c. none of the above

1.d. perpendicular

1.e. parallel

1.f. skew

2.a. never

2.b. always

2.c. always

2.d. sometimes, sometimes, sometimes

Step-by-step explanation:

For these questions, it is critical to know the definitions of each of the terms.

<h3>Definitions</h3>

Perpendicular: Two intersecting lines that form a right angle.
Parallel: Two non-intersecting lines that are coplanar.
Skew: Two non-intersecting lines that are not coplanar.

Problem Breakdown

1.a. skew $\overset{\longleftrightarrow}{AB} \text{ and } \overset{\longleftrightarrow}{EF}$

Point E is not on line AB. Any three non-linear points form a unique plane (left face), so A, B, and E are coplanar. Point F is not in plane ABE. Since line EF and line AB do not intersect and are not coplanar, they are skew.

1.b. parallel $\overset{\longleftrightarrow}{CD} \text{ and } \overset{\longleftrightarrow}{EH}$

Line CD is parallel to line AB, and line AB is parallel to line EH. Parallel lines have a sort of "transitive property of parallelism," so any pair of those three lines is coplanar and parallel to each other.

1.c. none of the above $\overset{\longleftrightarrow}{AC} \text{ and } \overset{\longleftrightarrow}{GA}$

Line AC and line GA share point A, necessarily intersecting at A. Therefore, line AC and line GA cannot be parallel or skew.

Any three points form a unique plane, so these two lines are coplanar, however, do they form a right angle? If the figure is cube-shaped, as depicted, then no.

Note that each line segment AC, AG, and CG are all the diagonal of a cube face. A cube has equal edge lengths, so each of those diagonals would be equal. Thus, triangle ACG is an equilateral triangle.

All equilateral triangles are equiangluar, and by the triangle sum theorem, the measures of the angles of a planar triangle must sum to 180°, forcing each angle to have a measure of 60° (not a right angle). So, lines AC and GA are not perpendicular. (If the shape were a rectangular prism, these lines still aren't perpendicular, but the proof isn't as neat, there is a lot to discuss, and a character limit)

"none of the above"

1.d. perpendicular $\overset{\longleftrightarrow}{DG} \text{ and } \overset{\longleftrightarrow}{HG}$

Line DG and line HG share point G, so they intersect. Both lines are the edges of a face, so they intersect at a right angle. Perpendicular

1.e. parallel $\overset{\longleftrightarrow}{AC} \text{ and } \overset{\longleftrightarrow}{FH}$

Line AC is a diagonal across the top face, and line FH is a diagonal across the bottom face. They are coplanar in a plane that cuts straight through the cube from top to bottom, but diagonally through those faces.

1.f. skew $\overset{\longleftrightarrow}{CD} \text{ and } \overset{\longleftrightarrow}{AG}$

Point A is not on line CD, and any three non-linear points form a unique plane (the top face), so C, D, and A are coplanar. Point G is not in plane ACD. Since line CD does not intersect and is not coplanar with line AG, by definition, the lines are skew.

2.a. Two lines on the top of a cube face are never skew

Two lines are on a top face are necessarily coplanar since they are both in the plane of the top face. By definition, skew lines are not coplanar. Therefore, these can never be skew.

2.b. Two parallel lines are always coplanar.

If two lines are parallel, by definition of parallel lines, they are coplanar.

2.c. Two perpendicular lines are always coplanar

If two lines AB and CD are perpendicular, they form a right angle. To form a right angle, they must intersect at the right angle's vertex (point P).

Note that A and B are unique points, so either A or B (or both) isn't P; similarly, at least one of C or D isn't P. Using P, and one point from each line that isn't P, those three points form a unique plane, necessarily containing both lines. Therefore, they must always be coplanar.

2.d. A line on the top face of a cube and a line on the right side face of the same cube are sometimes parallel, sometimes skew, and sometimes perpendicular

Consider each of the following cases:

Parallel: Consider line CD (top face), and line GF (right face). They are coplanar and don't intersect. By definition, parallel.
Skew: Consider line AD (top face), and line GF (right face). They aren't coplanar and don't intersect. By definition, skew.
Perpendicular: Consider line AD (top face), and line GD (right face). They do intersect at D, and form a right angle. By definition, perpendicular.
None of the above: Consider line AC (top face), and line GC (right face). These lines intersect at C, but as discussed in part 1.c, they form a 60° angle, not a right angle. By definition, "none of the above".

These 4 cases prove it is possible for a pair of top face/right face lines to be parallel, perpendicular, skew, or none of the above. So, those two lines are neither "always", nor "never", one of those choices. Therefore, they are each "sometimes" one of them.

8 0

2 years ago