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2 years ago

I am having trouble and want to make sure this is the correct answer

Mathematics

1 answer:

sp2606 [1]2 years ago

3 0

The answer is (2x-1) x (22+9x) simplified. (The x in the middle is for multiplication)

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Any 10th grader solve it 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

Helppppppppppppppppppp I’ll you brainlist if you don’t respond if you’re going to put something random I will report you !

qwelly [4]

Answer:

x = 0

y = -4

negative 4 cuz the line goes down, meaning it's negative

4 0

3 years ago

If 100 people drink 1.5 cups of coffee a day and the cost per cup is .40 how much would it cost to provide coffee for the whole

N76 [4]

Answer:

It would cost 15600 to provide coffee for the whole year.

Step-by-step explanation:

Given:

If 100 people drink 1.5 cups of coffee a day and the cost per cup is .40

Now, to find the cost to provide coffee for the whole year if you're place is open 5 days a week.

So, we have the unit rate, and we have to find the cost of 1.5 cup by multiplying 0.40.

1.5 cup of coffee cost $1.5\times 0.40=0.60.$

Thus, 1.5 cup cost 0.60.

If, 100 people drink 1.5 cups.

Then, total cost of 100 people drinking 1.5 cups of coffee = $100\times 0.60=60$

As, place is open 5 days a week.

So, to get the cost of coffee in 5 days of a week we multiply 5 by 60:

$5\times 60\\\\=300.$

Hence, the cost of coffee in a week is 300.

In 1 year there are 52 weeks.

Now, to get the cost for the whole year we multiply 52 by 300:

$52\times 300\\\\=15600.$

Therefore, it would cost 15600 to provide coffee for the whole year.

6 0

3 years ago

Read 2 more answers

Tell someone a story about "1 more and then 1 more

gayaneshka [121]

Once upon a time there was a guy named Jeff. he went to a new school where he moved. so he was all by himself with no friends at all.. so then he met this guy named Zac so it was 1+1=2. and they were bff so they lived happily ever after together being bffs

4 0

4 years ago

Find the average rate of change over the given interval f(x) = 2^x , [3,5]

Vlada [557]

Answer:

$y=x*2^x (3,5)+y$

Step-by-step explanation:

4 0

3 years ago