Let G be some point on the diagonal line away from point E.
Angle DEG represents angle 1.
We're given that angle DEF is a right angle which means it's 90 degrees. Angle DEG is some angle smaller than 90 degrees. By definition, that must mean angle 1 is acute. Any acute angle is smaller than 90 degrees. There's not much else to say other than this is just a definition problem.
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Extra side notes:
If angle 1 was a right angle, then that would mean angle GEF would have to be 0 degrees; however the diagram shows this isn't the case.
If angle 1 was obtuse, then there's no way we'd be able to fit it into angle DEF. In other words, there's no way to have an angle larger than 90 fit in a 90 degree angle.
Answer:
The elements of S-block elements are 14, so therefore to create an inequality model we declare S to be the unknown variable.
therefore since we have 14 elements we use an inclusive class interval from 1-14. the model is given as
1≤S≥14
Step-by-step explanation:
you have to declare variables before for the unknowns.
you also declare the class interval to be inclusive class interval or exclusive class interval.
Answer:
I think it would be 15
Step-by-step explanation:
Because if you split the number line into 6 different sections (with the 15 on the left side and the 16 on the right side) and shade/fill in 2 of those, you would be closer to 15.
You forgot to put the problem but choose answers that will divide evenly
Let the initial point of the vector be (x,y). Then the magnitude of the vector v can be written as:
The magnitude of vecor v is given to be 10. So we can write:
Now from the given options, we have to check which one satisfies the above equation. That point will be the initial point of the vector.
The point in option d, satisfies the equation.
Thus, the answer to this question is option D
