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The equation t^3=a^2 shows the relationship between a planet’s orbital period, T, and the planet’s mean distance from the sun, A

ankoles [38]

Answer:

2√2

Step-by-step explanation:

We can find the relationship of interest by solving the given equation for A, the mean distance.

<h3>Solve for A</h3>

$T^3=A^2\\\\A=\sqrt{T^3}=T\sqrt{T}\qquad\text{take the square root}$

<h3>Substitute values</h3>

The mean distance of planet X is found in terms of its period to be ...

$D_x=T_x\sqrt{T_x}$

The mean distance of planet Y can be found using the given relation ...

$T_y=2T_x\\\\D_y=T_y\sqrt{T_y}=2T_x\sqrt{2T_x}=(2\sqrt{2})T_x\sqrt{T_x}\\\\D_y=2\sqrt{2}\cdot D_x$

The mean distance of planet Y is increased from that of planet X by the factor ...

2√2

8 0

2 years ago

Find the value of each power?

sweet [91]

1=-8 l
4=100 l
7=? l
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5 0

4 years ago

Read 2 more answers

Factor completely <br>5x^2-9x-2

wariber [46]

Answer:

(5x+1)(x - 2)

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Factor completely x^8- 16.

Nataly_w [17]

Answer:

Step-by-step explanation:

So in this example we'll be using the difference of squares which essentially states that: $(a-b)(a+b)=a^2-b^2$ or another way to think of it would be: $a-b=(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b})$ . So in this example you'll notice both terms are perfect squares. in fact x^n is a perfect square as long as n is even. This is because if it's even it can be split into two groups evenly for example, in this case we have x^8. so the square root is x^4 because you can split this up into (x * x * x * x) * (x * x * x * x) = x^8. Two groups with equal value multiplying to get x^8, that's what the square root is. So using these we can rewrite the equation as:

$x^8-16 = (x^4-4)(x^4+4)$

Now in this case you'll notice the degree is still even (it's 4) and the 4 is also a perfect square, and it's a difference of squares in one of the factors, so it can further be rewritten:

$x^4-4 = (x^2-2)(x^2+2)$

So completely factored form is: $(x^2-2)(x^2+4)(x^4+4)$

I'm assuming that's considered completely factored but you can technically factor it further. While the identity difference of squares technically only applies to difference of squares, it can also be used on the sum of squares, but you need to use imaginary numbers. Because $x^2+4 = x^2-(-4)$ . and in this case a=x^2 and b=-4. So rewriting it as the difference of squares becomes: $x^4+4 = x^4 - (-4) = (x^2-\sqrt{-4})(x^2+\sqrt{-4}) = (x^2-2i)(x^2+2i)$ just something that might be useful in some cases.