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ale4655 [162]
3 years ago
15

Find the next term of the given sequence. 1.31, 2.54, 3.77, ... 4.90 5.00 5.10

Mathematics
2 answers:
vfiekz [6]3 years ago
8 0

It is A.P. with common difference = 2.54-1.31 = 1.23

So, next term would be 3.77+1.23 = 5.00

OPTION B IS YOUR ANSWER

givi [52]3 years ago
4 0

Answer:

Option B is correct.

the next term of the sequence is 5.00

Explanation:

Arithmetic progression(A.P) is a sequence of numbers in which the consecutive terms are formed by adding a constant quantity with the preceding term.


Common difference(d) is the constant quantity stated in the above definition of arithmetic progression. it is given by;

d=a_{n+1} - a_n for all n∈N


Given sequence:-  1.31, 2.54 , 3.77 , ___ ;

we have;

a_1 = 1.31 , a_2= 2.54 and  a_3 = 3.77

First find the common difference;

d =a_2 - a_1 = 2.54-1.31 = 1.23

or

d = =a_3-a_2 = 3.77-2.54 =1.23

Therefore, by the definition of arithmetic progression,

the given sequence is Arithmetic progression

Now, to find the fourth term of the sequence, add constant quantity (d) to the third term;

a_4 = 3.77+1.23 = 5.00

Therefore, in the given sequence next term will be 5.00.


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Match each set of vertices with the type of triangle they form.
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Answer:  The calculations are done below.


Step-by-step explanation:

(i) Let the vertices be A(2,0), B(3,2) and C(5,1). Then,

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Since, AB = BC and AB² + BC² = CA², so triangle ABC here will be an isosceles right-angled triangle.

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AB=\sqrt{(4-6)^2+(2-2)^2}=\sqrt{4}=2,\\\\BC=\sqrt{(6-5)^2+(2-3.73)^2}=\sqrt{14.3729},\\\\CA=\sqrt{(5-4)^2+(3.73-2)^2}=\sqrt{14.3729}.

Since, BC = CA, so the triangle ABC will be an isosceles triangle.

(iii) Let the vertices be A(-5,2), B(-4,4) and C(-2,2). Then,

AB=\sqrt{(-5+4)^2+(2-4)^2}=\sqrt{5},\\\\BC=\sqrt{(-4+2)^2+(4-2)^2}=\sqrt{8},\\\\CA=\sqrt{(-2+5)^2+(2-2)^2}=\sqrt{9}.

Since, AB ≠ BC ≠ CA, so this will be an acute scalene triangle, because all the angles are acute.

(iv) Let the vertices be A(-3,1), B(-3,4) and C(-1,1). Then,

AB=\sqrt{(-3+3)^2+(1-4)^2}=\sqrt{9}=3,\\\\BC=\sqrt{(-3+1)^2+(4-1)^2}=\sqrt{13},\\\\CA=\sqrt{(-1+3)^2+(1-1)^2}=\sqrt 4.

Since AB² + CA² = BC², so this will be a right angled triangle.

(v) Let the vertices be A(-4,2), B(-2,4) and C(-1,4). Then,

AB=\sqrt{(-4+2)^2+(2-4)^2}=\sqrt{8},\\\\BC=\sqrt{(-2+1)^2+(4-4)^2}=\sqrt{1}=1,\\\\CA=\sqrt{(-1+4)^2+(4-2)^2}=\sqrt{13}.

Since AB ≠ BC ≠ CA, and so this will be an obtuse scalene triangle, because one angle that is opposite to CA will be obtuse.

Thus, the match is done.

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4 years ago
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Aleksandr [31]

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Answer:

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Step-by-step explanation:

Bob's total earnings were ...

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