3√5
The distance between two points on an XY plane is calculated using the distance formula, which is employed in coordinate geometry or Euclidean geometry. The x-coordinate, often known as the abscissa, is a point's separation from the y-axis. The y-coordinate, often known as the ordinate, refers to a point's separation from the x-axis. A point on the x-axis has coordinates of the form (x, 0), and a point on the y-axis has coordinates of the form (0, y). We utilize the Pythagoras theorem in this case to determine the separation between any two points in a plane.
Distance formula = √ ( x₁ - x₂)² + ( y₁ - y₂)²
= √ 6² + 3²
=3√5
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Answer:
<h3><u>Let's</u><u> </u><u>understand the concept</u><u>:</u><u>-</u></h3>
Here angle B is 90°
So 
and 
Are right angled triangle
So we use Pythagoras thereon for solution
<h3><u>Required Answer</u><u>:</u><u>-</u></h3>
perpendicular=p=8cm
Hypontenuse =h =10cm
According to Pythagoras thereon
- Now in
Perpendicular=p=8cm
Base =b=15cm
- We need to find Hypontenuse =AD(x)
According to Pythagoras thereon
Answer:
0.032
Step-by-step explanation:
Divide each term by 0.032 and simplify
Part A
Answer: The common ratio is -2
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Explanation:
To get the common ratio r, we divide any term by the previous one
One example:
r = common ratio
r = (second term)/(first term)
r = (-2)/(1)
r = -2
Another example:
r = common ratio
r = (third term)/(second term)
r = (4)/(-2)
r = -2
and we get the same common ratio every time
Side Note: each term is multiplied by -2 to get the next term
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Part B
Answer:
The rule for the sequence is
a(n) = (-2)^(n-1)
where n starts at n = 1
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Explanation:
Recall that any geometric sequence has the nth term
a(n) = a*(r)^(n-1)
where the 'a' on the right side is the first term and r is the common ratio
The first term given to use is a = 1 and the common ratio found in part A above was r = -2
So,
a(n) = a*(r)^(n-1)
a(n) = 1*(-2)^(n-1)
a(n) = (-2)^(n-1)
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Part C
Answer: The next three terms are 16, -32, 64
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Explanation:
We can simply multiply each previous term by -2 to get the next term. Do this three times to generate the next three terms
-8*(-2) = 16
16*(-2) = -32
-32*(-2) = 64
showing that the next three terms are 16, -32, and 64
An alternative is to use the formula found in part B
Plug in n = 5 to find the fifth term
a(n) = (-2)^(n-1)
a(5) = (-2)^(5-1)
a(5) = (-2)^(4)
a(5) = 16 .... which matches with what we got earlier
Then plug in n = 6
a(n) = (-2)^(n-1)
a(6) = (-2)^(6-1)
a(6) = (-2)^(5)
a(6) = -32 .... which matches with what we got earlier
Then plug in n = 7
a(n) = (-2)^(n-1)
a(7) = (-2)^(7-1)
a(7) = (-2)^(6)
a(7) = 64 .... which matches with what we got earlier
while the second method takes a bit more work, its handy for when you want to find terms beyond the given sequence (eg: the 28th term)
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