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3 years ago

100x+2y =1 and also 100x -2y=1 what is the value of y ?

1 answer:

4 0

Yeah bro just take the 2 and subtract on both side so the y is left alone flip the 100 so now its 001 divide it by 1 and the answer is 007 like the spy.

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SCORPION-xisa [38]

Binomial means there are two terms, so c² − 16 is the answer
mononomial means there is only one term, so -8c is the answer
written in descending form means from the highest degree (exponent) to the lowest, 4x² − 3x + 9 is the answer for question 3.
Question 4: use the same method for question 3, the third term is -7x

Question 5: not sure what this is: 2x2y. Is it 2x²y or 2x2y?
you need to add up the exponent of x and y. In this case, the highest degree is y's 4th power plus x's 1st power, which makes a 5.

7 0

2 years ago

A wall has an area of 40 ft. It’s height is 8 feet. Which equation could you use to find its width. Answers 40- w = 8

RideAnS [48]

8w=40

You get area by doing length x width. In this case, the “length” is the height. So you would need to figure out what you multiply 8 by to get 40.

4 0

3 years ago

You received 1⁄3 pound of candy from your grandmother, 1⁄2 pound of candy from your sister, but your best friend ate 1⁄5 pound o

OverLord2011 [107]

Give them the same denominator.
19/30 is what you have left

3 0

3 years ago

Read 2 more answers

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

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

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

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

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

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

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