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gayaneshka [121]
3 years ago
14

Can someone help me?

Mathematics
2 answers:
krek1111 [17]3 years ago
8 0
I say..3x1= 3+1= 4 :-)
Olenka [21]3 years ago
6 0
1.5 x 2 = 3 + 1 = 4
I hope this helps you out.

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Find the unknown length in the right triangle 27cm and 13 cm
Phoenix [80]

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c^2 = (27)^2 - (13)^2

c^2 = 560

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In the figure below, ⎯⎯⎯⎯⎯⎯⎯⎯,⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯, and ⎯⎯⎯⎯⎯⎯⎯⎯⎯ are medians of △. If KC = 108 and OC = 2n+10, then n=
Nimfa-mama [501]

The question was incomplete. Below you will find the missing content.

In the figure below, BL, AM, and CK are medians of △ABC. If KC = 108 and OC = 2n + 10, then n=

The picture is attached below.

The value of n is 31.

The median is the line connecting the vertex and its midpoint on the opposite side of the triangle.

The intersection of all 3 medians of the triangle is called the centroid.

As we know centroid divides the median in the ratio of 2:1.

In the given picture,

Medians of triangle △ABC are AM, BL, and CK.

So the centroid of the triangle is O.

Given that KC= 108

As the centroid O divides the line KC in 2:1.

Let OC=2x

KO= X

As KC= KO+OC

⇒ 108= x+2x

⇒ 3x=108

⇒ x=108/3

⇒ x=36

Then OC= 2x= 2*36= 72

As given in the question OC= 2n+10

putting the value of OC in above equation

⇒ 72= 2n+10

⇒ 2n= 72-10

⇒ 2n=62

⇒ n=62/2

⇒n=31

Therefore the value of n is 31.

Learn more about the median of the triangle

here: brainly.com/question/2264495

#SPJ10

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2 years ago
A rival chocolate company is larger. They have 11 times as many workers.
scoundrel [369]

Answer:

13.4

Step-by-step explanation:

4 0
3 years ago
The nth term of a sequence is 3n^2- 1
Thepotemich [5.8K]

Answer: The number that belongs to both sequences is 26.

Step-by-step explanation:

We have two sequences, let's call one as A and the other as B.

The n-th term of sequence A is written as:

aₙ = 3*n^2 - 1

the nth term of sequence B is written as:

bₙ = 30 - n^2

We want to find a term that belongs to both sequences, (it can be for different integers, we can use n for sequence A and x for sequence B)

Then we want to find:

aₙ = bₓ

where n and x are integer numbers.

Then we will heave:

3*n^2 - 1 = 30 - x^2

To find the pair, we could isolate one of the variables, then input different integers in the other variable and see if the outcome is also an integer.

Let's isolate n.

3*n^2 = 30 - x^2 + 1

3*n^2 = 31 - x^2

n^2 = (31 - x^2)/3

n = √(  (31 - x^2)/3)

Now let's input different values for x, and see if the outcome is also an integer, notice that x is in a negative term inside a square root, then we have only a few values of x such that the equation can be true.

Then let's start with x = 1.

n(1) = √(  (31 - 1^2)/3) = √(30/3) = √10

We know that √10 is not an integer.

now with x = 2,

n(2) = √(  (31 - 2^2)/3)  = √( (31 - 4)/3) = √(27/3) = √9 = 3

then if x = 2, we have n = 3.

Both of them are integers, then we get:

a₂ = 3*(3)^2 - 1 = 27 - 1 = 26

b₃ = 30 - 2^2 = 30 - 4 = 26

The number that belongs to both sequences is 26.

5 0
3 years ago
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