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solniwko [45]
3 years ago
9

Question 3

Mathematics
1 answer:
alexira [117]3 years ago
7 0

Answer: when x = 7, y = 16

Step-by-step explanation: Here, we know that y varies inversely as x.

When we have two sets of inversely related coordinates, x₁ and y₁,

and x₂ and y₂, we can use the product rule, shown below,

to find the missing value.

<h2>x₁ y₁ = x₂ y₂</h2><h2 />

Here, we know that y = 14 when x = 8 so

one set of coordinates will be 8 and 14.

We want to know the value of y when x = 7.

So our other coordinates will be 7 and y.

So we have (8)(14) = (7)(y).

Simplifying, (8)(14) is equal to 112 and (7)(y) is equal to 7y.

So we have 112 = 7y.

Now we simply solve for y by dividing both sides of the equation by 7.

The 7's on the right side cancel out and

on the left, 112 divided by 7 is equal to 16.

So we have 16 = y.

So when x is equal to 7, y is equal to 16.

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Simplify the expression state any excluded values<br> 2a^2-4a+2<br> ---------------<br> 3a^2-3
chubhunter [2.5K]

Answer:

The simplified form is \dfrac{2(x-1)}{3(x+1)}.

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Step-by-step explanation:

Given:

The expression given is:

\dfrac{2a^2-4a+2}{3a^2-3}

Let us simplify the numerator and denominator separately.

The numerator is given as 2a^2-4a+2

2 is a common factor in all the three terms. So, we factor it out. This gives,

=2(a^2-2a+1)

Now, a^2-2a+1=(a-1)(a-1)

Therefore, the numerator becomes 2(a-1)(a-1)

The denominator is given as: 3a^2-3

Factoring out 3, we get

3(a^2-1)

Now, a^2-1 is of the form a^2-b^2=(a-b)(a+b)

So, a^2-1=(a-1)(a+1)

Therefore, the denominator becomes 3(a-1)(a+1)

Now, the given expression is simplified to:

\frac{2a^2-4a+2}{3a^2-3}=\frac{2(x-1)(x-1)}{3(x-1)(x+1)}

There is (x-1) in the numerator and denominator. We can cancel them only if x\ne1 as for x=1, the given expression is undefined.

Now, cancelling the like terms considering x\ne1, we get:

\dfrac{2a^2-4a+2}{3a^2-3}=\dfrac{2(x-1)}{3(x+1)}

Therefore, the simplified form is \dfrac{2(x-1)}{3(x+1)}

The simplification is true only if  x\ne1. So, x =1 is the excluded value for the given expression.

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