94 MINUS 68 EQUAL 26 DOLLARS MIN HAS
Answer:
You could just replace all the given possible values of k in the inequality and see which ones are solutions, but let's solve this in a more interesting way:
First, remember how the absolute value works:
IxI = x if x ≥ 0
IxI = -x if x ≤ 0
Then if we have something like:
IxI < B
We can rewrite this as
-B < x < B
Now let's answer the question, here we have the inequality:
I-k -2I < 18
Then we can rewrite this as:
-18 < (-k - 2) < 18
Now let's isolate k:
first, we can add 2 in the 3 parts of the inequality:
-18 + 2 < -k - 2 + 2 < 18 + 2
-16 < -k < 20
Now we can multiply all sides by -1, remember that this also changes the direction of the signs, then:
-1*-16 > -1*-k > -1*20
16 > k > -20
Then k can be any value between these two limits.
So the correct options (from the given ones) are:
k = -16
k = -8
k = 0
Answer:
Step-by-step explanation:
a) A square is a rectangle. True
Reason: Property of Rectangle: (i) Opposite sides are equal and parallel. (ii) Each angle is 90 (iii) Diagonals are equal and bisect each other.
A square hold all these properties. A square is a rectangle.
b) A polygon with 21 sides has 432 possible diagonals. False
Reason: Number of diagonals of a polygon =
n --> number of sides of a polygon.
= = 189 diagonals
c) All three angles in an Isosceles triangle are equal. False
Reason: In an isosceles triangle, two angles are equal.
If three angles are equal, then that is an equilateral triangle.
d) The measures of the exterior angles of a nonagon, a nine-sided figure, have a sum of 360°. True.
Reason: The sum of measures of the exterior angles of any polygon is 360
7.83 x 10^-7 (the *^* is where the power goes)
Answer:
A perfect square is a whole number that is the square of another whole number.
n*n = N
where n and N are whole numbers.
Now, "a perfect square ends with the same two digits".
This can be really trivial.
For example, if we take the number 10, and we square it, we will have:
10*10 = 100
The last two digits of 100 are zeros, so it ends with the same two digits.
Now, if now we take:
100*100 = 10,000
10,000 is also a perfect square, and the two last digits are zeros again.
So we can see a pattern here, we can go forever with this:
1,000^2 = 1,000,000
10,000^2 = 100,000,000
etc...
So we can find infinite perfect squares that end with the same two digits.