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2 years ago

What is the value of y in the solution to the system of equations?

Mathematics

1 answer:

Alex17521 [72]2 years ago

4 0

Answer:

y-value = 8

Step-by-step explanation:

1/3x + 1/4y = 1 is the same as 4/12x + 3/12y = 1

which is the same as (4x + 3y)/12 = 1

which is the same as 4x + 3y = 12

add both equations and the 'y' will cancel and you will have 6x = -18

x = -3

2(-3) - 3y = -30

-6 - 3y = -30

-3y = -24

y = 8

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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

2 years ago

I'm not sure how to do this. Help please?

OLEGan [10]

The area is 65 m squared

7 0

3 years ago

Choose the ratio equivalent to 10/9

satela [25.4K]

<h3>Answer:</h3>

30/27, 20/18...

<h3>Solution:</h3>

There are <em>infinitely many ratios equivalent to 10/9.</em>
Here are <em>some </em>of them:
30/27
20/18
100/90
60/54

Hope it helps.

Do comment if you have any query.

3 0

2 years ago

Calculate the distance between each given pair of points. Round your answer to the nearest tenth, if necessary.

Ksju [112]

The answer is 5

Here are the steps:

First off, we will be using the distance formula of
$d=\sqrt{(x_ {2}- x_{2}) ^{2}+( y_{2}- y_{1})^2 }$

So we have the ordered pairs of (3,1) and (6,5)

Once you plug them into the formula it should look like this:
$d= \sqrt{(6-3)^2+(5-1)^2}$

Now we do the math inside the parenthesis and end up with:
$d=\sqrt{3^2+4^2}$

Then you multiply by the power and simplify to get:
$d=\sqrt{25}$

And the $\sqrt{25}$ =5

So your answer is 5

8 0

2 years ago

On the beach boardwalk, there were 20 different places

Grace [21]

250 total places to get food this year

7 0

2 years ago