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Mekhanik [1.2K]
3 years ago
10

Hey i need help solving these and not just the answer i really would like to know how

Mathematics
1 answer:
Mars2501 [29]3 years ago
3 0
Hello,

1)
Let's put some name of vertex:
A with angle A=45°
B with angle B=90°
C The 3 vertex of the triangle of left.
AC=9 , y=AB=BC
With Pythagore, we can say: y²+y²=9²==>2y²=81===>y=9/√2=9√2/2

The other triangle is BDC: cos(60°)=x/y==>x=y*1/2=9√2/4

2)
Let's give some names:

A with angle A=60°
B with angle B=90
C the 3th vertex of the  left triangle.
AB=10, BC=y
tan 60°=y/10=sin(60°)/cos(60°)=(√3/2)/(1/2)=√3==>y=10√3

D is the 3 th vertex of the  right  triangle.
We use Pythagore: y²+y²=x²
x²=2*(10√3)²===>x=10*√6

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Given:

The sequence is 25, 20, 15, 10, 5.

To find:

The ninth term of the given sequence.

Solution:

We have,

25, 20, 15, 10, 5

It is an AP because the difference between two consecutive terms are same.

Here,

First term (a) = 25

Common difference (d) = 20-25

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The nth terms of an AP is

a_n=a+(n-1)d

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Putting a=25, n=9 and d=-5 to get the 9th term.

a_9=25+(9-1)(-5)

a_9=25+(8)(-5)

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a_9=-15

Therefore, the ninth term of the given sequence is -15.

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