Answer:
In order to find equation:
<u>Find slope</u>:
Then find equation using:
y - y1 = m(x -x1) where (x1, y1) are points, m is slope
Factor out the greatest common factor for 18x2 + 42x – 12.
1 answer:
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Three important properties of the diagonals of a rhombus that we need for this problem are:
1. the diagonals of a rhombus bisect each other
2. the diagonals form two perpendicular lines
3. the diagonals bisect the angles of the rhombus
First, we can let O be the point where the two diagonals intersect (as shown in the attached image). Using the properties listed above, we can conclude that ∠AOB is equal to 90° and ∠BAO = 60/2 = 30°.
Since a triangle's interior angles have a sum of 180°, then we have ∠ABO = 180 - 90 - 30 = 60°. This shows that the ΔAOB is a 30-60-90 triangle.
For a 30-60-90 triangle, the ratio of the sides facing the corresponding anges is 1:√3:2. So, since we know that AB = 10, we can compute for the rest of the sides.
Similarly, we have
Now, to find the lengths of the diagonals,
So, the lengths of the diagonals are 10 and 10√3.
Answer: 10 and 10√3 units
1. the diagonals of a rhombus bisect each other
2. the diagonals form two perpendicular lines
3. the diagonals bisect the angles of the rhombus
First, we can let O be the point where the two diagonals intersect (as shown in the attached image). Using the properties listed above, we can conclude that ∠AOB is equal to 90° and ∠BAO = 60/2 = 30°.
Since a triangle's interior angles have a sum of 180°, then we have ∠ABO = 180 - 90 - 30 = 60°. This shows that the ΔAOB is a 30-60-90 triangle.
For a 30-60-90 triangle, the ratio of the sides facing the corresponding anges is 1:√3:2. So, since we know that AB = 10, we can compute for the rest of the sides.
Similarly, we have
Now, to find the lengths of the diagonals,
So, the lengths of the diagonals are 10 and 10√3.
Answer: 10 and 10√3 units
Answer:
Step-by-step explanation:
In order to write the series using the summation notation, first we need to find the nth term of the sequence formed. The sequence generated by the series is an arithmetic sequence as shown;
4, 8, 12, 16, 20...80
The nth term of an arithmetic sequence is expressed as Tn = a +(n-1)d
a is the first term = 4
d is the common difference = 21-8 = 8-4 = 4
n is the number of terms
On substituting, Tn = 4+(n-1)4
Tn = 4+4n-4
Tn = 4n
The nth term of the series is 4n.
Since the last term is 80, L = 4n
80 = 4n
n = 80/4
n = 20
This shows that the total number of terms in the sequence is 20
According to the series given 4 + 8 + 12 + 16 + 20+ . . . + 80 , we are to take the sum of the first 20terms of the sequence. Using summation notation;
4 + 8 + 12 + 16 + 20+ . . . + 80 =