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3 years ago

Consider the sequence: 12, 17, 22, ... , ... , .... What is the 405th term of this sequence?

Mathematics

2 answers:

Mice21 [21]3 years ago

7 0

A_n= a₁+(n-1)d

a₁ first term

n terms

d distance between each value

a_n= 12+(405-1)(5)=2032

kompoz [17]3 years ago

7 0

The first term of the sequence is 12. You can also tell that you’re adding five after the first term.
12+5(x-1)=y
Since you’re not automatically adding it on the first term, you need to subtract one from x.
You could also do this.
7+5x=y
You would still get the same number.
Let’s plug 405 in for x in the equations.
12+5(405-1)
12+5(404)
12+2020
2032
Let’s see if it’s the same thing in the second equation.
7+5(405)
7+2025
2032
The 405th term is 2032.

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First we are going to draw the triangle using the given coordinates.
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Distance formula: $d= \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

Distance from point A to point B:
$d_{AB}= \sqrt{[6-(-9)]^2+(10-5)^2}$
$d_{AB}= \sqrt{(6+9)^2+(10-5)^2}$
$d_{AB}= \sqrt{(15)^2+(5)^2}$
$d_{AB}= \sqrt{225+25}$
$d_{AB}= \sqrt{250}$
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Distance from point A to point C:
$d_{AC}= \sqrt{[2-(-9)]^2+(-10-5)^2}$
$d_{AC}= \sqrt{(2+9)^2+(-10-5)^2}$
$d_{AC}= \sqrt{11^2+(-15)^2}$
$d_{AC}= \sqrt{121+225}$
$d_{AC}= \sqrt{346}$
$d_{AC}= 18.60$

Distance from point B from point C
$d_{BC}= \sqrt{(2-6)^2+(-10-10)^2}$
$d_{BC}= \sqrt{(-4)^2+(-20)^2}$
$d_{BC}= \sqrt{16+400}$
$d_{BC}= \sqrt{416}$
$d_{BC}=20.40$

Now, we are going to find the semi-perimeter of our triangle using the semi-perimeter formula:
$s= \frac{AB+AC+BC}{2}$
$s= \frac{15.81+18.60+20.40}{2}$
$s= \frac{54.81}{2}$
$s=27.41$

Finally, to find the area of our triangle, we are going to use Heron's formula:
$A= \sqrt{s(s-AB)(s-AC)(s-BC)}$
$A=\sqrt{27.41(27.41-15.81)(27.41-18.60)(27.41-20.40)}$
$A= \sqrt{27.41(11.6)(8.81)(7.01)}$
$A=140.13$

We can conclude that the perimeter of our triangle is 140.13 square units.

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