Here , we are provided with a table which shows 5 consecutive terms of an arithmetic sequence . But before solving further , let's recall that ;
The n'th term of a Arithmetic Sequence let's say it be
is given by ;
Where , <u>d</u> is the common difference
Now , here we are given with ;
We have to find the 2nd , 3rd and 4th term respectively ,
Now , by using the above formula , 5th term can be written as ;
Putting the values and transposing 1st term to RHS , we have ;
Now , as we got the common difference , so we can find out the missing terms now ;



Now



Also ,



Now , The given table can be written as ;

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brainly.com/question/26750175
65
1. you work in the parentheses and multiply 2.8 by 5 which is 14
2. you divide 14 by 7 which is 2
3. you follow PEMDAS and multiply 26 by what is in the parentheses (2) which is 52
4. you add 13 with 52 which is 65
hope this helped!!
Answer:
Step-by-step explanation:
B(2,10); D(6,2)
Midpoint(x1+x2/2, y1+y2/2) = M ( 2+6/2, 10+2/2) = M(8/2, 12/2) = M(4,6)
Rhombus all sides are equal.
AB = BC = CD =AD
distance = √(x2-x1)² + (y2- y1)²
As A lies on x-axis, it y-co ordinate = 0; Let its x-co ordinate be x
A(X,0)
AB = AD
√(2-x)² + (10-0)² = √(6-x)² + (2-0)²
√(2-x)² + (10)² = √(6-x)² + (2)²
√x² -4x +4 + 100 = √x²-12x+36 + 4
√x² -4x + 104 = √x²-12x+40
square both sides,
x² -4x + 104 = x²-12x+40
x² -4x - x²+ 12x = 40 - 104
8x = -64
x = -64/8
x = -8
A(-8,0)
Let C(a,b)
M is AC midpoint
(-8+a/2, 0 + b/2) = M(4,6)
(-8+a/2, b/2) = M(4,6)
Comparing;
-8+a/2 = 4 ; b/2 = 6
-8+a = 4*2 ; b = 6*2
-8+a = 8 ; b = 12
a = 8 +8
a = 16
Hence, C(16,12)
Answer:
landslide
Step-by-step explanation: