How would you describe a carton back and hold 1000 unit cubes that measures 1" x 1" x 1" by using a unit can you
1 answer:
You might be interested in
Your question can be quite confusing, but I think the gist of the question when paraphrased is: P<span>rove that the perpendiculars drawn from any point within the angle are equal if it lies on the angle bisector?
Please refer to the picture attached as a guide you through the steps of the proofs. First. construct any angle like </span>∠ABC. Next, construct an angle bisector. This is the line segment that starts from the vertex of an angle, and extends outwards such that it divides the angle into two equal parts. That would be line segment AD. Now, construct perpendicular line from the end of the angle bisector to the two other arms of the angle. This lines should form a right angle as denoted by the squares which means 90° angles. As you can see, you formed two triangles: ΔABD and ΔADC. They have congruent angles α and β as formed by the angle bisector. Then, the two right angles are also congruent. The common side AD is also congruent with respect to each of the triangles. Therefore, by Angle-Angle-Side or AAS postulate, the two triangles are congruent. That means that perpendiculars drawn from any point within the angle are equal when it lies on the angle bisector
Please refer to the picture attached as a guide you through the steps of the proofs. First. construct any angle like </span>∠ABC. Next, construct an angle bisector. This is the line segment that starts from the vertex of an angle, and extends outwards such that it divides the angle into two equal parts. That would be line segment AD. Now, construct perpendicular line from the end of the angle bisector to the two other arms of the angle. This lines should form a right angle as denoted by the squares which means 90° angles. As you can see, you formed two triangles: ΔABD and ΔADC. They have congruent angles α and β as formed by the angle bisector. Then, the two right angles are also congruent. The common side AD is also congruent with respect to each of the triangles. Therefore, by Angle-Angle-Side or AAS postulate, the two triangles are congruent. That means that perpendiculars drawn from any point within the angle are equal when it lies on the angle bisector
It looks like you have the domain confused for the range! You can think of the domain as the set of all "inputs" for a function (all of the x values which are allowed). In the given function, we have no explicit restrictions on the domain, and no situations like division by 0 or taking the square root of a negative number that would otherwise put limits on it, so our domain would simply be the set of all real numbers, R. Inequality notation doesn't really use ∞, so you could just put an R to represent the set. In set notation, we'd write
and in interval notation,
The <em>range</em>, on the other hand, is the set of all possible <em>outputs</em> of a function - here, it's the set of all values f(x) can be. In the case of quadratic equations (equations with an x² term), there will always be some minimum or maximum value limiting the range. Here, we see on the graph that the maximum value for f(x) is 3. The range of the function then includes all values less than or equal to 3. As in inequality, we can say that
,
in set notation:
(this just means "f(x) is a real number less than or equal to 3")
and in interval notation: