It might be helpful to draw a diagram of a line in this case, but you could do this algebraically as well. The midpoint of any segment is basically the average of the two endpoints of the segment. Thus, for example, the midpoint of AC would be the value of A added to the value of C, which is then divided by 2 to get B. You just have to find all of the values for each of the letters, and then find the distance between B and D, for instance, which (the distance) may be represented as a system of equations with the statement of the distance for AE.
I believe that the answer is 1/8. I hope this helps
Answer: this many 4
Step-by-step explanation:
2(2x+7)+2(3x+8y)
=4x+14+6x+16y
=10x+16y+14
Answer:
{-2, -1 , 3}
Step-by-step explanation:
When we have a set like:
{x₁, x₂, x₃}
The mode is the value that appears the most, so if there is no mode, then each value appears just one time.
The median is the middle value, here we know that the median is -1, then we can rewrite the set as:
{x₁, -1 , x₃}
The mean is computed as:
Mean = (x₁ + x₂ + x₃)/3
in this case we know that the mean is 0, then:
0 = (x₁ + x₂ + x₃)/3
then the numerator must be zero, so:
0 = (x₁ + x₂ + x₃)
replacing the value of x₂ = -1 we get:
0 = (x₁ - 1 + x₃)
where:
-5 < x₁ < -1 < x₃ ≤ 3
Now we can select the values of x₁ and x₃ such that the sum is equal to zero, and it meets the wanted restrictions.
here we can choose x₃ to be equal to 3 (the maximum allowed value), I do this because I noticed that the other values that are larger than -1 will not work (just with quick math).
then:
0 = x₁ - 1 + 3
Now we can solve this for x₁
0 = x₁ + 2
-2 = x₁
Then the set is:
{-2, -1 , 3}
Answer:
Simplify the denominator.
Tap for more steps...
x
(
x
+
3
)
(
x
−
3
)
⋅
3
x
x
2
−
5
x
+
6
Factor
x
2
−
5
x
+
6
using the AC method.
Tap for more steps...
x
(
x
+
3
)
(
x
−
3
)
⋅
3
x
(
x
−
3
)
(
x
−
2
)
Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.
(
x
+
3
)
(
x
−
3
)
,
(
x
−
3
)
(
x
−
2
)
The LCM is the smallest positive number that all of the numbers divide into evenly.
1. List the prime factors of each number.
2. Multiply each factor the greatest number of times it occurs in either number.
The number
1
is not a prime number because it only has one positive factor, which is itself.
Not prime
The LCM of
1
,
1
is the result of multiplying all prime factors the greatest number of times they occur in either number.
1
The factor for
x
+
3
is
x
+
3
itself.
(
x
+
3
)
=
x
+
3
(
x
+
3
)
occurs
1
time.
The factor for
x
−
3
is
x
−
3
itself.
(
x
−
3
)
=
x
−
3
(
x
−
3
)
occurs
1
time.
The factor for
x
−
2
is
x
−
2
itself.
(
x
−
2
)
=
x
−
2
(
x
−
2
)
occurs
1
time.
The LCM of
x
+
3
,
x
−
3
,
x
−
3
,
x
−
2
is the result of multiplying all factors the greatest number of times they occur in either term.
(
x
+
3
)
(
x
−
3
)
(
x
−
2
)
Step-by-step explanation:
there does that help
