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2 years ago

6/a - 4/b = 1 and 9/a - 8/b = 1

Mathematics

1 answer:

Alja [10]2 years ago

5 0

Answer:

a = 3 ; b = 4

Step-by-step explanation:

Eqn.1 = $\frac{6}{a} - \frac{4}{b} = 1$

Eqn.2 = $\frac{9}{a} - \frac{8}{b} = 1$

Both the eqns have RHS (Right Hand Side) equal. So,

$\frac{6}{a} - \frac{4}{b} = \frac{9}{a} - \frac{8}{b}$

$= > \frac{6b - 4a}{ab} = \frac{9b - 8a}{ab}$

Cancelling ab from both the denominators of both the sides,

$= > 6b - 4a = 9b - 8a$

$= > 8a - 4a = 9b - 6b$

$= > 4a = 3b$

$= > a = \frac{3b}{4}$

Putting the value of a in Eqn.1,

$\frac{6}{ \frac{3b}{4} } - \frac{4}{b} = 1$

$= > \frac{24}{3b} - \frac{4}{b} = 1$

$= > \frac{8}{b} - \frac{4}{b} = 1$

$= > \frac{8 - 4}{b} = 1$

$= > b = 4$

Putting the value of b in the value of a

$a = \frac{3 \times 4}{4} = 3$

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2 years ago

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Answer:

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Step-by-step explanation:

From the figure, the coordinate of the tail of the vector,

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By using the distance formula,

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2 years ago

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Suppose that f: R --> R is a continuous function such that f(x +y) = f(x)+ f(y) for all x, yER Prove that there exists KeR su

Pachacha [2.7K]

<h2>Answer with explanation:</h2>

It is given that:

f: R → R is a continuous function such that:

$f(x+y)=f(x)+f(y)------(1)$ ∀ x,y ∈ R

Now, let us assume f(1)=k

Also,

f(0)=0

( Since,

f(0)=f(0+0)

i.e.

f(0)=f(0)+f(0)

By using property (1)

Also,

f(0)=2f(0)

i.e.

2f(0)-f(0)=0

i.e.

f(0)=0 )

Also,

f(2)=f(1+1)

i.e.

f(2)=f(1)+f(1) ( By using property (1) )

i.e.

f(2)=2f(1)

i.e.

f(2)=2k

Similarly for any m ∈ N

f(m)=f(1+1+1+...+1)

i.e.

f(m)=f(1)+f(1)+f(1)+.......+f(1) (m times)

i.e.

f(m)=mf(1)

i.e.

f(m)=mk

Now,

$f(1)=f(\dfrac{1}{n}+\dfrac{1}{n}+.......+\dfrac{1}{n})=f(\dfrac{1}{n})+f(\dfrac{1}{n})+....+f(\dfrac{1}{n})\\\\\\i.e.\\\\\\f(\dfrac{1}{n}+\dfrac{1}{n}+.......+\dfrac{1}{n})=nf(\dfrac{1}{n})=f(1)=k\\\\\\i.e.\\\\\\f(\dfrac{1}{n})=k\cdot \dfrac{1}{n}$

Also,

when x∈ Q

i.e. $x=\dfrac{p}{q}$

Then,

$f(\dfrac{p}{q})=f(\dfrac{1}{q})+f(\dfrac{1}{q})+.....+f(\dfrac{1}{q})=pf(\dfrac{1}{q})\\\\i.e.\\\\f(\dfrac{p}{q})=p\dfrac{k}{q}\\\\i.e.\\\\f(\dfrac{p}{q})=k\dfrac{p}{q}\\\\i.e.\\\\f(x)=kx\ for\ all\ x\ belongs\ to\ Q$

(

Now, as we know that:

Q is dense in R.

so Э x∈ Q' such that Э a seq belonging to Q such that:

$\to x$ )

Now, we know that: Q'=R

This means that:

Э α ∈ R

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$a_n\ belongs\ to\ Q$

and

$a_n\to \alpha$

$f(a_n)=ka_n$

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This means that:

$f(a_n)\to f(\alpha)\\\\i.e.\\\\k\cdot a_n\to f(\alpha)\\\\Also\\\\k\cdot a_n\to k\alpha$

This means that:

$f(\alpha)=k\alpha$

This means that:

f(x)=kx for every x∈ R

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2 years ago

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sattari [20]

This system can be a bit tricky unless you write it out:
(2n + 2)(2n - 2) ~ First you have to take one variable and multiply it to each of the other two variables, like:
2n * 2n ~ And:
2n * -2 ~ This gives you:
4n^2 - 4n
Now we do this with the other:
2 * 2n
2 * -2
4n - 4
Now we combine them both while adding like terms:
4n^2 - 4
Therefore your answer is: A. 4n^2 - 4
(This is due to how when having the positive 4n subtracted by the negative it cancels it out. Leaving us with the remaining two terms left.)

I hope this helps, have a great rest of your day! ^ ^
~Ghostgate

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2 years ago

Rewrite the equation below so that it does not have fractions.

Eduardwww [97]

Answer:

6x-40=3

Step-by-step explanation:

3/4 x -5 =3/8

Multiply each side by 8 to get rid of the fractions.

8(3/4x -5 ) =8*3/8

Distribute

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2 years ago

Read 2 more answers

6/a - 4/b = 1 and 9/a - 8/b = 1​

6/a - 4/b = 1 and 9/a - 8/b = 1