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2 years ago

Whenever a,x,y are positive integers, which of the following expressions is equivalent to x^ay^a

Mathematics

1 answer:

lilavasa [31]2 years ago

3 0

Answer:

${(xy)}^{a}$

Step-by-step explanation:

when the exponent is equal, we can put the x and y together in a bracket and a as the Exponent of xy.

Hope I get your question~

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If the independent variable becomes very large what would be the expected change in the dependent variable

jenyasd209 [6]

Answer: The independent variable is the variable the experimenter changes or controls and is assumed to have a direct effect on the dependent variable. ... The dependent variable is the variable being tested and measured in an experiment, and is 'dependent' on the independent variable.

Step-by-step explanation:

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3 years ago

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 $\dfrac{2(x-1)}{3(x+1)}$ .

$x =1$ is the excluded value for the given expression.

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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3 years ago

PLZ HELP I NEED HELP PLZ

Llana [10]

Answer:

ok I will help

Step-by-step explanation:

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2 years ago

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3 years ago

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son4ous [18]

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