The sum of the first 20 terms of an arithmetic sequence with the 18th term of 8.1 and a common difference of 0.25 is 124.5
Given,
18th term of an arithmetic sequence = 8.1
Common difference = d = 0.25.
<h3>What is an arithmetic sequence?</h3>
The sequence in which the difference between the consecutive term is constant.
The nth term is denoted by:
a_n = a + ( n - 1 ) d
The sum of an arithmetic sequence:
S_n = n/2 [ 2a + ( n - 1 ) d ]
Find the 18th term of the sequence.
18th term = 8.1
d = 0.25
8.1 = a + ( 18 - 1 ) 0.25
8.1 = a + 17 x 0.25
8.1 = a + 4.25
a = 8.1 - 4.25
a = 3.85
Find the sum of 20 terms.
S_20 = 20 / 2 [ 2 x 3.85 + ( 20 - 1 ) 0.25 ]
= 10 [ 7.7 + 19 x 0.25 ]
= 10 [ 7.7 + 4.75 ]
= 10 x 12.45
= 124.5
Thus the sum of the first 20 terms of an arithmetic sequence with the 18th term of 8.1 and a common difference of 0.25 is 124.5
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Answer: Exact Form: x = 7/2
Decimal Form: x = 3.5
Mixed Number Form: x = 3 1/2
How to: <u>
Solve for x by simplifying both sides of the equation, then isolating the variable.</u>
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In the given situation, the ratio of the perimeters of two rectangles is 4 to 7. The perimeter of the larger rectangle is 42 inches.Now, let's find the perimeter of the smaller.Let's solve for the answerRatio and formula:=> 4 : 7 = X = 42, where x is the unknown perimeter of the smaller rectangle=> 4 * 42 = 168 / 7 = 24Thus the answer is letter D. 24 inches is the parameter of smaller rectangle.<span>
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Answer:
there are 11 big-bags and 24 small-bags.
Step-by-step explanation:
there are x big-bags and y small-bags.
so now we can know:
(1) x + y = 35
(2) 12x + 7y = 300
in (1),we can do like this:
both left and right x7
then (x + y)x7 = 35 x 7
then 7x + 7y = 35 x 7
then 7x + 7y = 245
now,
(1)7x + 7y =245
(2)12x + 7y = 300
we can both left and right do this: (2) - (1)
then
(12x + 7y) - (7x + 7y) = 300 -245
then
12x + 7y - 7x - 7y = 300 -245
then
12x - 7x +7y - 7y =300 -245
then
5x =55
then
5x ÷ 5 = 55 ÷ 5
then
x = 11
because
x + y =35;x=11
so
11 + y =35
11+ y -11 = 35 -11
then
y = 24
now we know:there are 11 big-bags and 24 small-bags.
*see attachment for the figure described
Answer:
5 units
Step-by-step explanation:
==>Given the figure attached below, let where FH and EG intercepted be K.
Since FH are midpoints of parallel lines, KE = KG = x.
Given that the area of the kite EFGH = 35 square units, and we know the length of one of the diagonals = HF = KF + KH = 2 + 5 = 7, we can solve for x using the formula for the area of a kite.
Area of kite = ½ × d1 × d2
Where d1 = KH = 7
d2 = EG = KE + KG = x + x = 2x
Area of kite EFGH = 35
THUS:
35 = ½ × 7 × 2x
35 = 1 × 7 × x
35 = 7x
Divide both sides by 7
35/7 = x
x = 5
