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3 years ago

Which value of x makes the numerical sentence true 25^1/2 x 5^-3=5x

1 answer:

4 0

$25^{\frac{1}{2}} \times 5^{-3} = 5^x$

$(5^2)^{\frac{1}{2}} \times 5^{-3} = 5^x$

$5^1 \times 5^{-3} = 5^x$

$5^{1 - 3} = 5^x$

$5^{-2} = 5^x$

$x = -2$

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The equation 7^2=a^2 shows the relationship between a planet’s orbital period, T, and the planet’s mean distance from the sun, A

jeyben [28]

Answer:

The period of Y increases by a factor of $k^ {3/2}$ with respect to the period of X

Step-by-step explanation:

The equation $T ^ 2 = a ^ 3$ shows the relationship between the orbital period of a planet, T, and the average distance from the planet to the sun, A, in astronomical units, AU. If planet Y is k times the average distance from the sun as planet X, at what factor does the orbital period increase?

For the planet Y:

$T_y ^ 2 = a_y ^ 3$

For planet X:

$T_x ^ 2 = a_x ^ 3$

To know the factor of aumeto we compared $T_x$ with $T_y$

We know that the distance "a" from planet Y is k times larger than the distance from planet X to the sun. So:

$a_y ^ 3 = (a_xk) ^ 3$

$\frac{T_y ^ 2}{T_x ^ 2}=\frac{a_y ^ 3}{a_x^ 3}\\\\\frac{T_y ^ 2}{T_x ^ 2}=\frac{(a_xk)^3}{a_x ^ 3}\\\\\frac{T_y^ 2}{T_x^ 2}=\frac{k ^ {3}a_{x}^ 3}{a_{x}^ 3}\\\\\frac{T_{y}^ 2}{T_{x}^ 2}=k ^ 3\\\\T_{y}^ 2 = T_{x}^{2}k^{3}\\\\T_{y} =k^{\frac{3}{2}}T_x$

Then, the period of Y increases by a factor of $k^ {3/2}$ with respect to the period of X

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3 years ago

Find the slope of the line that is defined by 2y=3x-30

Elden [556K]

Answer: 1.5

Step-by-step explanation:

The equation of a line in slope - intercept form is given as :

y = mx + c , where :

m = slope

c = y - intercept

Given:

2y = 3x - 30

What we need to do is to write it in slope - intercept form, we will make y the subject of formula.

Divide the equation through by 2

2y/2 = 3x/2 - 30/2

y = 3x/2 - 15

comparing with y = mx + c

slope = 3/2 = 1.5

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3 years ago

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Luba_88 [7]

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the exact answer is 20..

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sattari [20]

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