Ellipses, parabolas, and infinity. A parabola can be defined as the locus of points equidistant from a fixed point (called the focus) and a fixed line (called the directrix). But we know from projective geometry that parabolas and ellipses are essentially the same object.
For this, simply use the rules of logs:
- Adding logs of the same base (which is 7 here) is the same as multiplying
-Subtracting logs of the same base is dividing
So the single log would be:
log₇(x*y/z) which is the same as
log₇(xy/z)
For the given figure, x = 34.640 units.
Step-by-step explanation:
Step 1:
The formulae needed to solve this question are:
The triangle containing x can be split into two right-angled triangles.
So the hypotenuse of the triangle containing the length x and the already given right angle triangle's hypotenuse is the same length.
Step 2:
First, we calculate the hypotenuse of the given triangle.
The angle of the triangle is 45°, the adjacent side's length = 10√3 units. Assume the hypotenuse's length to be a units.
units.
Step 3:
For the split triangle, we have the angle as 45°, the opposite side measuring units (only half of x is in each triangle) and the hypotenuse measuring 24.498 units.
So x measures 34.640 units.
Answer:
Yes, the difference of two rational numbers is a rational number. The reason for this lies in the following facts: The product of two integers is an integer. The difference between two integers is an integer.
Step-by-step explanation:
Answer:
1 < x < 4 . . . . {x | x < 4 <u>and</u> x > 1}
Step-by-step explanation:
We want to write the answer as a compound inequality, if possible. As it is written, we can solve each separately.
x + 1 < 5
x < 4 . . . . . . . subtract 1
__
x -4 > -3
x > 1 . . . . . . . add 4
So, the solution is ...
(x < 4) ∩ (x > 1) . . . . . . the intersection of the two solutions
As a compound inequality, this is written ...
1 < x < 4
_____
<em>Comment on the problem</em>
The two answer choices shown don't make any sense. You might want to have your teacher demonstrate the solution to this problem.