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ollegr [7]
3 years ago
13

Can you help me ASAP please?

Mathematics
2 answers:
siniylev [52]3 years ago
7 0
It is iii , if you need me to explain it to you I will message your account otherwise it's iii
serg [7]3 years ago
4 0
I think it's (3) Calculate the nth term rule of the sequence (I think the answer to the equation is 8xy+36x+6y+9)
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Any 10th grader solve it <br>for 50 points​
kkurt [141]

Answer:

\frac{a}{p}\times (q-r)+\frac{b}{q}\times (r-p)+\frac{c}{r}\times (p-q)\neq 0  is proved for the sum of pth, qth and rth terms of an arithmetic progression are a, b,and c respectively.

Step-by-step explanation:

Given that the sum of pth, qth and rth terms of an arithmetic progression are a, b and c respectively.

First term of given arithmetic progression is A

and common difference is D

ie., a_{1}=A and common difference=D

The nth term can be written as

a_{n}=A+(n-1)D

pth term of given arithmetic progression is a

a_{p}=A+(p-1)D=a

qth term of given arithmetic progression is b

a_{q}=A+(q-1)D=b and

rth term of given arithmetic progression is c

a_{r}=A+(r-1)D=c

We have to prove that

\frac{a}{p}\times (q-r)+\frac{b}{q}\times (r-p)+\frac{c}{r}\times (p-q)=0

Now to prove LHS=RHS

Now take LHS

\frac{a}{p}\times (q-r)+\frac{b}{q}\times (r-p)+\frac{c}{r}\times (p-q)

=\frac{A+(p-1)D}{p}\times (q-r)+\frac{A+(q-1)D}{q}\times (r-p)+\frac{A+(r-1)D}{r}\times (p-q)

=\frac{A+pD-D}{p}\times (q-r)+\frac{A+qD-D}{q}\times (r-p)+\frac{A+rD-D}{r}\times (p-q)

=\frac{Aq+pqD-Dq-Ar-prD+rD}{p}+\frac{Ar+rqD-Dr-Ap-pqD+pD}{q}+\frac{Ap+prD-Dp-Aq-qrD+qD}{r}

=\frac{[Aq+pqD-Dq-Ar-prD+rD]\times qr+[Ar+rqD-Dr-Ap-pqD+pD]\times pr+[Ap+prD-Dp-Aq-qrD+qD]\times pq}{pqr}

=\frac{Arq^{2}+pq^{2} rD-Dq^{2} r-Aqr^{2}-pqr^{2} D+qr^{2} D+Apr^{2}+pr^{2} qD-pDr^{2} -Ap^{2}r-p^{2} rqD+p^{2} rD+Ap^{2} q+p^{2} qrD-Dp^{2} q-Aq^{2} p-q^{2} prD+q^{2}pD}{pqr}

=\frac{Arq^{2}-Dq^{2}r-Aqr^{2}+qr^{2}D+Apr^{2}-pDr^{2}-Ap^{2}r+p^{2}rD+Ap^{2}q-Dp^{2}q-Aq^{2}p+q^{2}pD}{pqr}

=\frac{Arq^{2}-Dq^{2}r-Aqr^{2}+qr^{2}D+Apr^{2} -pDr^{2}-Ap^{2}r+p^{2}rD+Ap^{2}q-Dp^{2}q-Aq^{2}p+q^{2}pD}{pqr}

\neq 0

ie., RHS\neq 0

Therefore LHS\neq RHS

ie.,\frac{a}{p}\times (q-r)+\frac{b}{q}\times (r-p)+\frac{c}{r}\times (p-q)\neq 0  

Hence proved

5 0
3 years ago
Click photo for question
Lana71 [14]

In a triangle when an angle between two equal length sides is 60 that means that the triangle is equilateral.
6 0
3 years ago
In the data set below, what is the mean absolute deviation?
lbvjy [14]

Add the numbers and divide by 5 to get a mean of 6.4

8 - 6.4 = 1.6

4 - 6.4 = -2.4

7 - 6.4 = 0.6

8 - 6.4 = 1.6

5 - 6.4 = -1.4

Add the absolute value of each: 1.6 + 2.4 + 0.6 + 1.6 + 1.4 = 7.6

Divide by 5 = 1.52

1.5

3 0
2 years ago
What is the answer to 7x-7=12x+13
QveST [7]

Answer:

x  =  − 4

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
Can please cross multiply this for me :)
Alchen [17]
64 x 100 = 6400
11 x x = 11x
I’ll solve this out for you
11x = 6400
Divide 11 from both sides
x = 581.8
Hope this helps!
4 0
3 years ago
Read 2 more answers
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